23 Definitions

8. Angle

An angle is formed when two rays meet at a point.

Notation: ∠ABC.

9. Rectilinear Angle

An angle formed by two straight lines.

10. Right Angle

When a line meets another so that the two adjacent angles are equal, each is a right angle.

Notation: ∠ABC = 90° (π/2 rad).

11. Obtuse Angle

Greater than a right angle.

Notation: ∠ABC > 90°.

12. Acute Angle

Less than a right angle.

Notation: ∠ABC < 90°.

13. Boundary

The limit of a figure.

14. Figure

A shape enclosed by boundaries.

15. Circle

The set of all points at a fixed distance (radius) from a center O.

Notation: c(O,r) is the circle centered at O with radius r, that is, all points A such that |OA| = r.

16. Center of Circle

The point O from which all radii are equal.

17. Diameter

A line through the center, ending on the circle.

Notation: AB with O midpoint; length = 2r.

18. Semicircle

Half a circle, bounded by a diameter and the arc it cuts off.

Notation: arc(AB).

19. Rectilinear Figure

A figure bounded by straight lines (a polygon).

20. Triangle

A polygon with three sides.

Notation: △ABC.

21. Quadrilateral

A polygon with four sides.

Notation: □ABCD.

22. Polygon

A rectilinear figure with more than four sides.

23. Types of Triangles

By sides: equilateral (three equal), isosceles (two equal), scalene (none equal).
By angles: right (one right), obtuse (one obtuse), acute (all acute).

Notation: equilateral |AB| = |BC| = |CA|; isosceles |AB| = |AC|; right ∠ABC = 90°, etc.

5 Postulates

1. Line between Points

A straight line can be drawn joining any two points.

Notation: segment AB.

2. Extension of Line

A finite line can be extended indefinitely.

Notation: extend AB to C.

3. Circle

A circle can be drawn with any center and radius.

Notation: c(O,r).

4. Right Angles

All right angles are equal.

Notation: if ∠ABC = 90° and ∠DEF = 90°, then they are equal.

5. Parallel Postulate

Given a line and a point not on it, exactly one line can be drawn through the point parallel to the given line.

Notation: through P not on ℓ, there is a unique m with m ∥ ℓ.

5 Common Notions

1. Equality (Transitive)

Things equal to the same thing are equal to one another.

Notation: if a = b and b = c, then a = c.

2. Addition Rule

If equals are added to equals, the results are equal.

Notation: if a = b, then a + c = b + c.

3. Subtraction Rule

If equals are subtracted from equals, the remainders are equal.

Notation: if a = b, then a − c = b − c.

4. Coincidence Principle

Things which coincide with one another are equal to one another.

Notation: if △ABC ≅ △DEF, then all corresponding parts equal.

5. Whole vs. Part

The whole is greater than the part.

Notation: if B lies between A and C, then |AB| < |AC|.

Equilateral Triangle

An equilateral triangle can be constructed on any given finite straight line.

ElementsI.1 Take a stick and mark its ends A and B. With your compass on A, swing a circle through B. Then with your compass on B, swing a circle through A. Where the circles meet, call that point C. Join A to C and B to C. You’ve made a triangle with all three sides equal.

Proof

By the definition of a circle, every point on c_A is at distance |AB| from A; in particular, |AC| = |AB|. Likewise, every point on c_B is at distance |AB| from B; in particular, |BC| = |AB|. Hence |AC| = |BC| = |AB|, so the sides of △ABC are equal and the triangle is equilateral. Q.E.D.

Construction

1. Let AB be the given segment.
2. Draw circle c_A centered at A with radius |AB|.
3. Draw circle c_B centered at B with radius |AB|.
4. Let C be a point of intersection: C = cA ∩ cB, with C ≠ A,B.
5. Join AC and BC to form △ABC.

Triangle Congruences

Side–Angle–Side (SAS)

If two triangles have two pairs of corresponding sides equal and the included angles equal, then the triangles are congruent.

ElementsI.4 Make a triangle by hinging two sticks at a fixed angle. If a second triangle uses sticks of the same lengths and the same hinge angle, the two triangles match exactly when you lay one on top of the other.

Proof

Let △ABC and △DEF satisfy |AB| = |DE|, |AC| = |DF|, and ∠BAC = ∠EDF. Superpose △ABC onto △DEF by placing A onto D and aligning AB with DE. Because the included angles at A and D are equal, ray AC coincides with ray DF, and since |AC| = |DF|, point C falls on F. Thus △ABC and △DEF coincide: △ABC ≅ △DEF. Corresponding bases and angles are equal. Q.E.D.

Side–Side–Side (SSS)

If two triangles have all three pairs of corresponding sides equal, then the triangles are congruent.

ElementsI.8 Build two triangles from the same three stick lengths. There’s only one way (up to flipping) to join them: the triangles match.

Proof

Let △ABC and △DEF satisfy |AB| = |DE|, |BC| = |EF|, |CA| = |FD|. Place A on D with AB along DE. The locus of points at distance |CA| from A is a circle; likewise the locus at distance |FD| from D is the same radius circle. The equalities force the third vertex to the same intersection point on the same side of the base, so C falls on F. Hence △ABC ≅ △DEF. Corresponding angles are equal. Q.E.D.

Angle–Side–Angle (ASA) and AAS

If two triangles have two angles equal respectively and one corresponding side equal (either the included side or a side opposite one of the equal angles), then the triangles are congruent.

ElementsI.26 Hinge two angles the same way and make one side the same; the third vertex lands in only one place, so the triangles match.

Proof

Let △ABC and △DEF satisfy ∠A = ∠D, ∠B = ∠E and either |AB| = |DE| (adjacent case) or |AC| = |DF| (opposite case). Equal two angles force the third angles equal. In the adjacent case, the included angle between the given sides is equal; Side-Angle-Side applies. In the opposite case, placing the equal side and opening the equal adjacent angles determines the remaining side uniquely, yielding superposition of the triangles. Thus △ABC ≅ △DEF. Q.E.D.

Isosceles Triangle

Base Angles

In any triangle with two equal sides, the angles at the base are equal.

ElementsI.5 Make a triangle where the two sides from the top point are the same length—like two equal sticks hinged at the top. When you set the base on the table, the “feet” sit symmetrically. The tilt at the left foot matches the tilt at the right foot.

Proof

Let △ABC satisfy |AB| = |AC|. Compare △ABC with △ACB (swap the labels of B and C). Then the side pairings are |AB| ↔ |AC| and |AC| ↔ |AB| (equal by the given), and the included angle is ∠BAC in both triangles. By SAS congruence, △ABC ≅ △ACB. Therefore the base angles correspond and are equal: ∠ABC = ∠ACB. Q.E.D.

Converse: Opposite Sides

In any triangle, if two base angles are equal, then the opposite sides are equal.

ElementsI.6If a triangle “leans” the same amount at both feet, then the two side sticks up to the top must be the same length—otherwise one side would reach higher or lower and spoil the symmetry.

Proof

Let △ABC satisfy ∠ABC = ∠ACB. Compare △ABC with △ACB (swap B and C). They have two equal angles (∠A is common; ∠B = ∠C by the given) and the corresponding side |BC| = |CB|. By ASA/AAS congruence, △ABC ≅ △ACB. Therefore the sides opposite the equal base angles are equal: |AB| = |AC|. Q.E.D.

Bisection

Segment Bisection

Any finite straight line can be cut into two equal parts.

ElementsI.10 Take a stick AB. Swing equal-radius arcs from A and B that cross above and below the stick. The line through the crossings meets AB at its midpoint M; it’s also ⟂ to AB.

Proof

Let circles with the same radius meet at X and Y. Then |XA| = |XB| and |YA| = |YB|. In △XAB and △YAB, equal radii give two equal sides; the base angles at A and B match, so XY is the perpendicular bisector of AB. Hence the intersection M = XY ∩ AB satisfies |AM| = |MB|. Q.E.D.

Construction

1. Let AB be given.
2. Draw circle c_A centered at A with any radius > |AB|/2.
3. Draw circle c_B centered at B with the same radius.
4. Let X, Y be the two intersections: {X,Y} = cA ∩ cB.
5. Draw the line XY; let M = XY ∩ AB. Then |AM| = |MB| and XY ⟂ AB.

Angle Bisection

Any rectilinear angle can be cut into two equal angles.

ElementsI.9 From the angle’s vertex A, swing a circle to mark points E on AB and F on AC. With the same opening, swing arcs from E and F to meet at D. The ray AD bisects the angle.

Proof

With |AE| = |AF| and equal radii from E and F, triangles △AED and △AFD have two sides equal and the included side AD common, so by SAS they are congruent. Hence ∠EAD = ∠DAF. Q.E.D.

Construction

1. Given rays AB and AC with common endpoint A.
2. Draw circle c_A centered at A to meet the rays at E (on AB) and F (on AC).
3. With the same radius, draw circles c_E and c_F.
4. Let D be an intersection of c_E and c_F inside the angle.
5. Draw AD; then ∠EAD = ∠DAC.

Perpendiculars

Perpendicular at a Point on a Line

From a point on a given straight line, a straight line can be drawn at right angles to the given line.

ElementsI.11 On the line ℓ, pick equal points B and C on either side of A. With the same radius, swing arcs from B and C to meet at X and Y. The line XY passes through A and is ⟂ to ℓ.

Proof

From the equal-radius circles at B and C, we have |XB|=|XC| and |YB|=|YC|. Thus X and Y lie on the perpendicular bisector of BC. Since A is the midpoint of BC, the line through X and Y is ⟂ to ℓ at A. Q.E.D.

Perpendicular from a Point to a Line

From a point not on a given straight line, a straight line can be drawn perpendicular to the given line.

ElementsI.12 With center P, swing a circle to meet the line at X and Y. Bisect XY at M. The line PM is ⟂ to the base line.

Proof

The points X and Y are equidistant from P. The midpoint M of XY lies on the perpendicular bisector of XY. Since P is also equidistant from X and Y, the line PM is that perpendicular bisector, hence ⟂ to the base line at M. Q.E.D.

Exterior Angle Theorem

In any triangle, if a side is produced, the exterior angle is greater than either of the two non-adjacent interior angles.

ElementsI.16 For △ABC, extend side BC to D. Then the exterior angle at C satisfies ∠ACD > ∠A and ∠ACD > ∠B.

Proof

Bisect AC at E (1.4). Join BE, and produce it beyond E to F so that BE = EF. Join FC. Then AE = CE, BE = EF, and the vertical angles at E are equal (I.15), so △ABE ≅ △CEF (SAS, I.4). Hence ∠A = ∠ECF. Since ray CF lies inside the exterior angle ∠ACD, we have ∠ECF < ∠ACD, so ∠A < ∠ACD. Repeating the same construction with the midpoint of BC shows ∠B < ∠ACD. Q.E.D.

Parallel Through a Point

Through a given point, a straight line can be drawn parallel to a given line.

ElementsI.31 Given a line ℓ and a point P not on it, draw any transversal PA to the line. At A the transversal makes an angle with ℓ. Copy that same angle at P on the opposite side of the transversal (using the angle-copy construction). The new ray through P is parallel to ℓ.

Proof

The transversal PA cuts the two lines ℓ and the new line through P, creating equal alternate interior angles by construction. By I.27, when a transversal makes equal alternate interior angles with two lines, the lines are parallel. Q.E.D.

Alternate Interior Angles

A transversal falling on two parallel lines makes the alternate interior angles equal, and the co-interior angles supplementary.

ElementsI.29 Let two parallel lines be cut by a transversal. The alternate interior angles — the pair on opposite sides of the transversal, between the parallels — are equal. The co-interior angles — the pair on the same side — sum to two right angles.

Proof

Suppose the alternate angles are not equal, and (say) ∠1 > ∠2. Add ∠3 (the supplement of ∠2 on the same side) to both: ∠1 + ∠3 > ∠2 + ∠3 = two right angles. But by Postulate 5, lines whose interior angles sum to less than two right angles meet on that side — contradicting the assumption that the lines are parallel. Therefore ∠1 = ∠2, and since ∠2 + ∠3 is a straight angle, ∠1 + ∠3 = two right angles. Q.E.D.

Sum of Angles in a Triangle

In any triangle, the sum of the three interior angles equals two right angles.

ElementsI.32 The three angles inside every triangle add up to exactly a straight angle — two right angles, or 180°. The proof extends one side of the triangle and draws a parallel through the opposite vertex, then uses alternate interior and corresponding angles to rearrange the three angles into a straight line.

Proof

In △ABC, extend side BC to D. Through C, draw CE parallel to BA (I.31). Since BA ∥ CE and AC is a transversal, the alternate interior angles are equal: ∠BAC = ∠ACE (I.29). Since BA ∥ CE and BD is a transversal, the corresponding angles are equal: ∠ABC = ∠ECD. Therefore ∠ACB + ∠ACE + ∠ECD = ∠ACB + ∠BAC + ∠ABC. The left side is the straight angle BCD, which equals two right angles. Hence the three angles of the triangle sum to two right angles. Q.E.D.

Triangle Inequality

In any triangle, the sum of any two sides is greater than the remaining side.

ElementsI.20 Pick up any triangle and measure its three sides. Add the two shorter ones together—their sum always exceeds the longest side. If it didn't, the two short sticks couldn't reach each other to close the figure. This is the fundamental constraint on when three lengths can form a triangle.

Proof

Let △ABC be given. Extend BA beyond A to a point D such that |DA| = |AC|. Then △DAC is isosceles, so ∠ACD = ∠ADC (I.5). Now ∠BCD = ∠BCA + ∠ACD > ∠ACD = ∠ADC = ∠BDC. In △BDC, the side opposite the larger angle is longer, so |BD| > |BC|. But |BD| = |BA| + |AD| = |BA| + |AC|. Therefore |BA| + |AC| > |BC|. The same argument applies to every pair of sides. Q.E.D.

Triangle From Three Lengths

Given three straight lines such that any two are together greater than the third, to construct a triangle with sides equal to them.

ElementsI.22 Given three lengths where any two exceed the third (I.20), lay down one as the base. Open the compass to the second length and sweep an arc from one endpoint; open it to the third length and sweep an arc from the other endpoint. The two arcs meet at the third vertex of the required triangle.

Proof

The vertex C lies on the circle of radius b centred at A, so |AC| = b. It also lies on the circle of radius c centred at B, so |BC| = c. The base |AB| = a by construction. The triangle inequality guarantees the circles intersect. Q.E.D.

Parallelograms

Properties

In a parallelogram the opposite sides and angles are equal, and the diagonal bisects the area.

ElementsI.34 Draw diagonal AC of parallelogram ABCD. Since AB ∥ DC, the alternate interior angles ∠BAC = ∠DCA (I.29). Likewise AD ∥ BC gives ∠DAC = ∠BCA. The diagonal AC is common, so △ABC ≅ △CDA by ASA (I.26). Therefore AB = DC, AD = BC, ∠DAB = ∠BCD, and ∠ABC = ∠CDA. Q.E.D.

Equal Area

Parallelograms on the same base and between the same parallels are equal in area.

ElementsI.35 Let ABCD and EBCF be parallelograms sharing base BC and lying between the same two parallel lines. Since AD = BC = EF (opposite sides of parallelograms), and AD ∥ EF, the figure ADFE is itself a parallelogram. Adding △ABE to both parallelograms or subtracting the common overlap shows the two original parallelograms have equal area.

ElementsI.36 The same conclusion holds when the bases are merely equal (not identical) and the parallelograms lie between the same parallels, since joining the equal bases produces a larger parallelogram and the argument of I.35 applies to each half.

Triangle–Parallelogram Area Relation

If a parallelogram and a triangle share the same base and lie between the same parallels, the parallelogram is double the triangle.

ElementsI.41 Place a parallelogram and a triangle on the same base between two parallel lines. The parallelogram’s diagonal splits it into two triangles, each equal in area to the given triangle (since they share the same base and height). So the parallelogram is exactly twice the triangle.

Proof

Let parallelogram ABCD and △EBC share base BC and lie between parallels through A,D,E and through B,C. Draw diagonal AC. By I.34, △ABC = △ACD, so the parallelogram = 2 × △ABC. Since △ABC and △EBC share base BC and lie between the same parallels, they are equal in area (a consequence of I.35 and I.34). Therefore parallelogram ABCD = 2 × △EBC. Q.E.D.

Equal-Area Constructions

To construct a parallelogram equal in area to a given triangle, in a given angle.

ElementsI.42 Given △ABC, bisect BC at E. Through A draw AF parallel to EC, and through C draw CF parallel to AE. The parallelogram AECF has the same area as the original triangle.

Proof

Since E is the midpoint of BC, △ABE = △AEC (equal bases, same height). Diagonal AE bisects parallelogram AECF (I.34), so the parallelogram = 2 × △AEC = 2 × ½ △ABC = △ABC. Q.E.D.

ElementsI.45 This generalizes to any rectilinear figure: decompose it into triangles, apply I.42 to each, and combine the resulting parallelograms along a common side. The final parallelogram has the same total area and the desired angle.

The Pythagorean Theorem

In right-angled triangles, the square on the side opposite the right angle equals the sum of the squares on the other two sides.

ElementsI.47 This is the crown jewel of Book I. Given a right triangle, build a square on each side. Drop an altitude from the right-angle vertex to the hypotenuse and extend it across the large square, splitting it into two rectangles. Each rectangle equals one of the smaller squares.

Proof

Let △ABC have a right angle at C. Draw squares on all three sides. Drop the altitude from C perpendicular to AB, meeting it at H, and extend through the square on AB. Consider the rectangle on the A-side of the altitude. Since △CAB and the square on CA share base CA and lie between the same parallels, the square = 2 × △CAB-half (I.41). By a matching argument the rectangle = 2 × the same triangle. So the rectangle equals the square on CA. Likewise the other rectangle equals the square on CB. The two rectangles together form the square on AB. Therefore the square on AB = the square on CA + the square on CB. Q.E.D.

Converse

If the square on one side of a triangle equals the sum of the squares on the other two sides, the angle opposite the first side is a right angle.

ElementsI.48 Given △ABC with AB² = AC² + BC², construct a right triangle DEF with DE = AC, EF = BC, and a right angle at E. By I.47, DF² = DE² + EF² = AC² + BC² = AB², so DF = AB. The two triangles are congruent by SSS (I.8), so ∠C = ∠E = a right angle. Q.E.D.

Algebraic Identity

If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.

ElementsII.4 Let the line have length a + b. Build the square on the whole line. One pair of opposite corners gives squares of side a and side b; the two remaining rectangles each measure a × b. Adding the four pieces recovers the familiar identity (a + b)² = a² + 2ab + b².

Proof

Draw the square on AB where AB = a + b. Let P be the point on AB at distance a from A. Through P draw a line parallel to the sides, and through the corresponding point on the opposite side draw another, dividing the square into four regions. The top-left and bottom-right are squares of sides a and b. The top-right and bottom-left are congruent rectangles of sides a and b. Therefore the square on the whole equals the two squares plus twice the rectangle. Q.E.D.

Difference of Two Squares

If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.

ElementsII.5 Take line AC bisected at M and cut again at D (between M and C). Set a = MC (the half) and b = MD. Then AD = a + b and DC = a − b. The proposition says AD · DC + MD² = MC², i.e. (a + b)(a − b) + b² = a², which gives the difference-of-squares identity (a + b)(a − b) = a² − b².

Proof

Draw the square on MC (side a). Inside it, mark the small square on MD (side b) at one corner. The remaining L-shaped gnomon has area a² − b². But this gnomon can also be rearranged into a rectangle of sides a + b and a − b, equalling AD · DC. Therefore AD · DC = a² − b². Q.E.D.

Geometric Quadratic Equation

If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.

ElementsII.6 Line AB is bisected at M, and segment BD is added in the same line. Set a = MB (the half) and b = BD (the extension). Then MD = a + b and AD = 2a + b. The proposition says AD · DB + MB² = MD², i.e. (2a + b) · b + a² = (a + b)². Expanding confirms the identity: 2ab + b² + a² = a² + 2ab + b².

This is Euclid’s method of completing the square. Given a rectangle of known area (the gnomon AD · DB), adding the square on the half converts it into a perfect square — exactly the technique used to solve quadratic equations.

Proof

Draw the square on MD (side a + b). Inside it, mark the square on MB (side a) at one corner. The L-shaped gnomon surrounding it has area MD² − MB² = (a + b)² − a² = 2ab + b² = (2a + b) · b = AD · DB. Therefore AD · DB + MB² = MD². Q.E.D.

The Golden Ratio

To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

ElementsII.11 This is the construction of the golden section. Given AB, find a point H on it such that AB · HB = AH². The ratio AB : AH equals the golden ratio φ = (1 + √5) / 2 ≈ 1.618.

The construction proceeds by building the square on AB, bisecting one side at E, and swinging an arc of radius EB to find a point F on the extension of that side. The length AF is then transferred back onto AB to give the golden section point H. The golden ratio will return in Book IV for constructing the regular pentagon.

Construction

On AB construct square ABDC. Let E bisect AC. Then EB = √(AB² + AE²) = √(s² + (s/2)²) = s√5 / 2. Extend CA beyond A to F with EF = EB. Then AF = EF − EA = s(√5 − 1)/2. With center A and radius AF, mark H on AB. By II.6 applied to line CF (bisected at E, with A on the segment), CF · FA + AE² = EF² = EB² = AE² + AB², so CF · FA = AB². Now CF = CA + AF = AB + AH and FA = AH, so (AB + AH) · AH = AB², which rearranges to AB · HB = AH². Q.E.D.

The Prototypical Law of Cosines

In obtuse-angled triangles, the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle and the projection of the other side upon it.

ElementsII.12 Let △ABC have an obtuse angle at B. Drop the altitude from C to line BA (extended beyond B), meeting it at D. Then AC² = AB² + BC² + 2 · AB · BD. This is the geometric form of the law of cosines for obtuse angles: the extra term 2 · AB · BD plays the role of −2ab cos C (which is positive when the angle is obtuse).

In acute-angled triangles, the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle and the projection of the other side upon it.

ElementsII.13 Now let △ABC be acute at B. The altitude from C meets BA at a point D between B and A. Then AC² = AB² + BC² − 2 · AB · BD. Together with II.12, these two propositions express the law of cosines: c² = a² + b² − 2ab cos C, the sign of the correction depending on whether the angle is obtuse or acute. The Pythagorean theorem (I.47) is the special case where the angle is right and the correction vanishes.

Equal-area Square

To construct a square equal to a given rectilinear figure.

ElementsII.14 Any polygon can first be converted to a rectangle of equal area (by I.45). This proposition then converts a rectangle into a square of the same area. Given a rectangle with sides p and q, we must find a length x = √(pq) so that x² = pq.

The key insight uses a semicircle. Place p and q end-to-end along a line, draw a semicircle on the total length, and erect a perpendicular from the junction. Where the perpendicular meets the semicircle, its height is exactly √(pq). This is a consequence of Thales’ theorem (Book III): the angle in the semicircle is a right angle, and the altitude of that right triangle creates two smaller triangles whose proportionality gives the geometric mean.

Construction

Lay segment AB = p and BC = q in a straight line. Let M be the midpoint of AC. Draw a semicircle with centre M and radius MA. Erect a perpendicular at B, meeting the semicircle at P. Then BP² = AB · BC = pq, and the square on BP has the same area as the rectangle. Q.E.D.

The Center

To find the centre of a given circle.

ElementsIII.1 A circle is defined as the set of all points equidistant from a single point — its centre. But if all you are given is the curve itself, how do you recover that hidden point? Euclid’s answer uses a fact we already know from I.10: the perpendicular bisector of any segment passes through every point equidistant from its endpoints. Since the centre is equidistant from any two points on the circle, it lies on the perpendicular bisector of every chord. Two such bisectors fix the centre uniquely.

Construction

Draw any chord AB. Bisect it at M and erect the perpendicular at M — this line passes through the centre. Draw a second chord CD, bisect it at N, and erect its perpendicular. The two perpendiculars meet at the centre O. Q.E.D.

The Tangent

From a given point to draw a straight line touching a given circle.

ElementsIII.17 A tangent is a line that touches a circle at exactly one point. Given an external point P and a circle with centre O, Euclid constructs the tangent by an elegant trick: find the midpoint M of OP and draw a circle centred at M passing through both O and P. This auxiliary circle meets the given circle at the tangent point T. Because the angle OTP is inscribed in a semicircle of the auxiliary circle, it is a right angle (III.31), so OT ⊥ PT — exactly the condition for a tangent.

Construction

Join OP and bisect it at M. With centre M and radius MO, draw a circle. Let T be an intersection of this circle with the given circle. Then PT is tangent to the given circle at T, because OT is a radius and the angle in a semicircle is right. Q.E.D.

Angles in Circles

In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base.

ElementsIII.20 An inscribed angle is formed by two chords that meet at a point on the circle. A central angle is formed by two radii meeting at the centre. Proposition III.20 reveals their relationship: if both subtend the same arc, the central angle is exactly twice the inscribed angle. The proof draws the radius from the centre to the inscribed-angle vertex, creating two isosceles triangles (since all radii are equal, by I.5), whose base angles combine to give the result.

Proof

Let A be on the circle and O the centre. Draw AO and extend it to the far side of the circle. Triangles OAB and OAC are isosceles (radii OA = OB = OC). By the exterior-angle theorem (I.16), each central sub-angle equals twice the corresponding base angle at A. Adding them gives ∠BOC = 2 ∠BAC. Q.E.D.

ElementsIII.21 An immediate corollary: two inscribed angles that subtend the same arc are equal, because each is half the same central angle.

ElementsIII.22 For a quadrilateral inscribed in a circle (a cyclic quadrilateral), opposite angles subtend arcs that together make up the whole circle. Since the full central angle is 360°, each pair of opposite inscribed angles sums to half of that — 180°.

Equal Angles and Equal Arcs

In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.

ElementsIII.26 This proposition and its converse (III.27) establish a fundamental correspondence: in circles of equal radius, a central angle uniquely determines an arc, and an arc uniquely determines a central angle. The proof is by superposition — place one circle atop the other so the centres and one radius coincide. If the central angles are equal, the second radii also coincide, and therefore the arcs between them coincide.

Proof

Let two equal circles have centres O₁ and O₂, with equal central angles. Place circle 2 on circle 1 so that O₂ falls on O₁ and one radius of each angle aligns. Because the angles are equal, the other radii also align. The circles are equal, so the arcs between the aligned radii coincide. Conversely, if the arcs are equal, the chords are equal (III.29), and the triangles formed by the radii and chords are congruent by I.8 (SSS), giving equal angles. Q.E.D.

Thales’ Theorem

In a circle the angle in the semicircle is right.

ElementsIII.31 This celebrated result — traditionally attributed to Thales of Miletus — is a special case of III.20. A diameter subtends a “central angle” of 180°, so any inscribed angle that subtends the same arc is half of that: 90°. We used this fact already in the tangent construction (III.17) and foreshadowed it when constructing the equal-area square (II.14). Now we prove it directly.

Proof

Let AB be a diameter of a circle with centre O, and let C be any point on the circle other than A or B. Draw the radius OC. Since OA = OC (radii), triangle OAC is isosceles, so ∠OAC = ∠OCA. Likewise OB = OC gives ∠OBC = ∠OCB. The angles of triangle ABC sum to 180° (I.32): ∠A + ∠B + ∠C = 180°. But ∠C = ∠OCA + ∠OCB = ∠A + ∠B, so 2 ∠C = 180° and ∠C = 90°. Q.E.D.

Intersecting Chords

If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

ElementsIII.35 When two chords cross inside a circle at a point E, the products of their segments are equal: AE · EB = CE · ED. The proof uses similar triangles — the inscribed angles ∠AEC and ∠DEB subtend arcs that force triangles AEC and DEB to be similar (by III.21 and vertical angles). Equal ratios of corresponding sides give the product equality. This is the first glimpse of the power of a point — the product AE · EB depends only on the distance from E to the centre, not on the particular chord.

Proof

Let chords AB and CD meet at E. Join AC and BD (mentally). Inscribed angles ∠CAB and ∠CDB subtend the same arc BC, so they are equal (III.21). Angles ∠AEC and ∠DEB are vertical angles. Therefore triangles AEC and DEB are similar (AA), giving AE/DE = CE/EB, hence AE · EB = CE · ED. Q.E.D.

The Power of a Point

If from a point without a circle there fall on it two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the part of it outside the circle will be equal to the square on the tangent.

ElementsIII.36 The intersecting-chords theorem (III.35) works when the crossing point lies inside the circle. Propositions III.36 and III.37 handle the case when the point lies outside. If two secants are drawn from an external point P, cutting the circle at A, B and C, D respectively, then PA · PB = PC · PD. In modern language, all three theorems (III.35, III.36, III.37) express the same invariant: the power of the point with respect to the circle, equal to d² − r² where d is the distance from the point to the centre.

Proof (III.36)

From external point P draw two secants. The inscribed angles ∠PAC and ∠PDB subtend the same arc, making triangles PAC and PDB similar (AA, sharing angle at P). Hence PA/PD = PC/PB, giving PA · PB = PC · PD. Q.E.D.

ElementsIII.37 The tangent is the limiting case of a secant: both intersection points coalesce into the single tangent point T. The product PA · PB becomes PT · PT = PT², giving the elegant formula PT² = PA · PB. This is consistent with the tangent construction: the right angle at T makes PT the geometric mean of PA and PB.

Inscribed Circles

To inscribe a circle in a given triangle; to circumscribe a circle about a given triangle.

ElementsIV.4 Every triangle has an incircle — a circle that fits snugly inside, tangent to all three sides. Its centre, called the incentre, is the point where the three angle bisectors meet (we already know how to bisect an angle from I.9). Drop a perpendicular from the incentre to any side; its length is the inradius, and the circle with that radius touches every side.

Construction

Bisect angles A and B. The bisectors meet at the incentre I. Drop a perpendicular from I to side BC, meeting it at D. Draw a circle centred at I with radius |ID|. This circle is tangent to all three sides, because the incentre is equidistant from every side. Q.E.F.

ElementsIV.5 Every triangle also has a circumcircle — a circle passing through all three vertices. Its centre, the circumcentre, lies at the intersection of the perpendicular bisectors of the sides (we know how to construct these from I.10 and I.11). In III.1 we already found the centre of a given circle using exactly the same idea; here we construct the circle from scratch.

Construction

Bisect sides AB and BC at their midpoints and erect perpendiculars there. The perpendiculars meet at the circumcentre O. Draw a circle centred at O through A. Since |OA| = |OB| = |OC| (the circumcentre is equidistant from all three vertices), the circle passes through all three vertices. Q.E.F.

Triangle in a Circle

To construct an isosceles triangle having each of the base angles double the remaining one.

ElementsIV.10 This is the 36°–72°–72° triangle — the building block of the regular pentagon. Its secret is the golden ratio: cut a leg at the point D where the short piece equals the base, and the ratio of the whole leg to the base is φ = (1 + √5) / 2. We showed in II.11 how to find this golden section; now we put it to work.

Proof

Let ABC be isosceles with |AB| = |AC| and ∠A = 36°, so that ∠B = ∠C = 72° (since the angles sum to 180°). Mark the point D on AB where |BD| = |BC|. By the golden-ratio construction, |AB| / |BD| = φ. Triangle BDC is isosceles with base angles equal to 36° at D and C, confirming that each base angle of the original triangle is double the apex angle. Q.E.D.

Pentagon in a Circle

To inscribe an equilateral and equiangular pentagon in a given circle.

ElementsIV.11 With the 36°–72°–72° triangle in hand (from IV.10), the side of a regular pentagon inscribed in a circle can be computed. The construction is elegant: fix one vertex on the circle, then step around with a compass set to the side length. Five equal chords divide the circle into five equal arcs.

The diagonal of a regular pentagon is to its side as the golden ratio φ ≈ 1.618 — the same ratio that appeared in the triangle of IV.10 and in II.11.

Construction

Given a circle with centre O, mark a point A₁ on the circle. Set your compass to the pentagon side length (derived from IV.10) and, pivoting at A₁, mark the intersection A₂ on the circle. Repeat from each new vertex until all five vertices are marked. Join consecutive vertices to form the pentagon. Q.E.F.

Hexagon in a Circle

To inscribe an equilateral and equiangular hexagon in a given circle.

ElementsIV.15 The regular hexagon is the simplest polygon to inscribe in a circle. Its key property is that the side length equals the radius — so no golden ratio is needed, just the same compass opening that drew the circle in the first place.

This is the same insight behind the equilateral triangle of I.1: two overlapping circles of equal radius produce an equilateral triangle, and six such triangles tile the interior of a circle exactly.

Construction

Given a circle with centre O and radius r, mark a point B₁ on the circle. Without changing the compass opening, pivot at B₁ and mark the intersection B₂ on the circle. Repeat from each new vertex. After five steps, the sixth vertex B₆ coincides with the starting point, confirming closure. Join consecutive vertices to form the hexagon. Q.E.F.

15-gon in a Circle

To inscribe an equilateral and equiangular fifteen-angled figure in a given circle.

ElementsIV.16 The final proposition of Book IV combines the pentagon (IV.11) and the hexagon (IV.15) into a single construction. A pentagon vertex spans an arc of 72°; a hexagon vertex spans 60°. The gap between a pentagon vertex and the next hexagon vertex is therefore 72° − 60° = 12°. Bisecting that gap gives 24° arcs, which is exactly 360° / 15. Repeating the 24° step fifteen times around the circle produces the regular 15-gon.

Proof

Inscribe a regular pentagon and a regular hexagon in the same circle, sharing a common vertex at the top. The pentagon’s next vertex is at 72° from the top; the hexagon’s next vertex is at 60°. The arc between them subtends 72° − 60° = 12°. Bisect that arc (by I.9 or III.26) to obtain two 12° arcs, giving adjacent vertices 24° apart. Since 360° / 24° = 15, stepping off this 24° arc repeatedly produces all fifteen vertices. Q.E.F.

Limit of Distance

No visible object is seen all at once; rather, it is seen part by part in succession. Yet by the quickness of sight we think we see it all at once.

OpticsProp. 3 For every visible object there is a distance beyond which it can no longer be seen. Visual rays diverge from the eye, so the farther an object lies, the fewer rays reach it and the smaller the angle it subtends. Eventually that angle falls below any perceptible threshold, and the object vanishes from sight.

Proof

Let the eye be at E and let AB be a visible segment. The visual rays from E to A and from E to B enclose the angle ∠AEB. As AB recedes from E, the angle ∠AEB diminishes continuously. Since the rays diverge, there is a distance at which ∠AEB becomes less than any given angle, and the object ceases to be seen. Q.E.D.

Apparent Size and Foreshortening

Of equal magnitudes situated at different distances from the eye, those nearer appear larger; and those farther appear smaller.

OpticsProp. 4 Two objects of the same true length look different when one is nearer. The nearer object subtends a wider angle at the eye and therefore appears larger. This is the geometric basis of all foreshortening.

Proof

Let AB and CD be equal segments, with AB nearer to the eye E. The visual rays from E to A and B enclose the angle ∠AEB, and the rays to C and D enclose ∠CED. Since AB is nearer and equal in length to CD, the angle ∠AEB > ∠CED, so AB appears larger. Q.E.D.

Of unequal magnitudes situated at the same distance from the eye, the greater appears greater and the lesser appears lesser.

OpticsProp. 5 Conversely, if two segments lie at the same distance but one is longer, the longer one subtends a wider angle and looks bigger. This is the complement of Prop. 4: apparent size depends on both true size and distance.

Proof

Let AB and CD lie at equal distance from the eye E, with AB > CD. Since they share the same distance but AB spans a wider interval, ∠AEB > ∠CED. Q.E.D.

Parallel Convergence

Parallel lines, when seen from a distance, appear to converge.

OpticsProp. 6 This is perhaps the most familiar optical phenomenon: railway tracks, colonnades, and corridors all seem to narrow in the distance. The explanation is immediate from the preceding propositions. The gap between the parallels is a constant width, but the angle it subtends at the eye shrinks with distance (5.2), so the lines appear to draw together.

Proof

Let AC and BD be parallel lines, and let the eye be at E. The cross-section AB near the eye subtends the angle ∠AEB, and the cross-section CD farther away subtends ∠CED. Since AB = CD (the lines are parallel and we compare equal transversals) but CD is more distant, by Prop. 4 we have ∠CED < ∠AEB. The parallels therefore appear closer together at greater distance. Q.E.D.

If an eye is placed between parallel lines, those to the left appear to converge to the right, and those to the right appear to converge to the left.

OpticsProp. 13 Standing in a corridor, both ends narrow equally. The same convergence principle applies in every direction from the eye.

Proof

Place the eye E between the parallel lines AC and BD. Looking left, the gap AB at distance subtends a diminishing angle; looking right, the gap CD at distance does the same. In both directions the parallels appear to converge, by the same reasoning as Prop. 6. Q.E.D.

Planes Rise and Fall to Eye Level

Planes lying below the eye appear to rise in the distance, and planes lying above the eye appear to descend, both tending toward the level of the eye.

OpticsProp. 11 & 12 A floor receding into the distance appears to slope upward; a ceiling appears to slope downward. Both converge on the horizon — the horizontal plane through the eye. This follows directly from the convergence of parallels (5.3): the floor and ceiling are parallel planes, and their visible edges subtend ever-smaller angles as they recede.

Proof (below the eye)

Let the eye be at E, elevated above a ground plane. The near edge of the ground at A makes a steep downward ray; the far edge at B makes a ray nearly horizontal. Successive points along the ground are seen at progressively higher angles, so the ground appears to rise toward the eye's level. Q.E.D.

Proof (above the eye)

Symmetrically, let a ceiling lie above the eye at E. The near edge at A makes a steep upward ray; the far edge at B makes a ray nearly horizontal. The ceiling appears to fall toward eye level. Q.E.D.

Circle or Ellipse?

Wheels of chariots sometimes appear circular and sometimes appear pressed together.

OpticsProp. 42 A circle viewed from the side looks like an ellipse. One diameter (perpendicular to the line of sight) keeps its true angular size, while the other (along the line of sight) is foreshortened. When both diameters subtend the same angle — that is, when you face the circle squarely — it looks round. At any other angle it looks compressed.

Proof

Let a circle with center O have a vertical diameter AB and a horizontal diameter CD. The eye at E views the circle obliquely. Since AB is perpendicular to the line of sight, it subtends its full angle. But CD is foreshortened: its apparent length is reduced because the near edge C and far edge D subtend a smaller angle than A and B at the same distance from O. Therefore ∠AEB > ∠CED, and the circle appears pressed into an ellipse. Q.E.D.

Perspective Depth

Of equal intervals on a straight line, those that are seen at a greater distance appear smaller.

OpticsProp. 50 A row of equal tiles receding across a floor looks compressed in the distance. Each tile subtends a smaller angle than the one before it, so the far tiles look shorter than the near ones. This is the geometric origin of depth compression in perspective drawing.

Proof

Let equal intervals AB, BC, CD lie along a line receding from the eye E. Each successive interval is farther from E, so by Prop. 4, ∠AEB > ∠BEC > ∠CED. The intervals appear progressively shorter. Q.E.D.

Size Increases Near Approach

If the eye approaches a visible object, the object appears to grow; if the eye recedes, the object appears to shrink.

OpticsProp. 57 This is the converse of the limit-of-distance idea (5.1). As you walk toward an object the visual angle widens, so the object fills more of your field of view. The effect is continuous: every step closer enlarges the apparent size.

Proof

Let AB be a fixed object. From the farther eye position E₁, the visual angle is ∠AE₁B. From the nearer position E₂, the visual angle is ∠AE₂B. Since E₂ is closer, ∠AE₂B > ∠AE₁B by Prop. 4, so the object appears larger from E₂. Q.E.D.

Motion Parallax

When the eye moves, nearer objects appear to move against the direction of the eye, and farther objects appear to move with it.

OpticsProp. 58 Look out a moving train window: nearby fences streak backward while distant mountains drift forward. Euclid explains this by comparing the angular shift of near and far objects when the viewpoint changes. The near object sweeps through a larger angle and so appears to move more.

Proof

Let the eye move from E₁ to E₂. A near object at A is seen at different angles from the two positions; a far object at B shifts by a smaller angle. The near object's apparent displacement is greater, creating the illusion that it moves opposite to the eye while the far object drifts along. Q.E.D.

Objects moving toward the eye appear to grow larger continuously.

OpticsProp. 59 This is the kinematic counterpart of Prop. 57. Whether the eye moves toward the object or the object moves toward the eye, the visual angle increases continuously, so the object appears to swell as it approaches.

Proof

Let an object of fixed size move from position A₁B₁ (far) to position A₂B₂ (near), while the eye E is fixed. Then ∠A₂EB₂ > ∠A₁EB₁, so the object appears larger in the nearer position. Q.E.D.

Relative Motion

If the eye is at the center of a circle, and an object moves along the circumference, those parts nearer the diameter through the eye appear to move faster.

OpticsProp. 55 Imagine standing at the center of a merry-go-round. An object on the rim moves at constant speed, but it appears to slow down near the top of its arc (directly ahead) and speed up near the sides (crossing your line of sight). The reason is angular velocity: near the top of the arc the motion is nearly radial (toward or away from you), changing the angle slowly; near the sides the motion is nearly tangential, sweeping the angle quickly.

Proof

Let the eye be at the center O of a semicircle with diameter AB. An object moves equal arcs in equal times. Near P (the top), the chord of the arc is nearly radial, so the visual ray turns very little. Near Q or R (the sides), the chord is nearly perpendicular to the ray, so the visual ray sweeps quickly. Equal arcs produce unequal angular displacements; the object appears faster at the sides. Q.E.D.

Measuring Height, Depth, and Length

The propositions above are theoretical. The final group in the Optics turns practical: how to determine an inaccessible height, a depth, or a length using nothing but shadows, a mirror, or a sighting rod. Each method reduces to similar triangles and the properties of visual rays.

Height by Shadow

To find the height of a tower by means of its shadow.

OpticsProp. 19 Plant a stick of known height in the ground and measure its shadow. At the same time, measure the tower's shadow. The sun's rays are parallel, so the stick and tower form similar right triangles. The ratio of the stick's height to its shadow equals the ratio of the tower's height to its shadow.

Proof

Let stick AB (height h) cast shadow AC (length s), and let tower DE (height H) cast shadow DF (length S). The sun's rays are parallel, so ∠BAC = ∠EDF. Both triangles are right-angled at the base, so they are similar. Hence H / S = h / s, giving H = h · S / s. Q.E.D.

Height by Mirror

To find the height of a tower by means of a mirror placed on the ground.

OpticsProp. 20 Place a mirror flat on the ground between you and the tower. Step back until you see the top of the tower in the mirror. The angle of incidence equals the angle of reflection, so the triangle from your eye to the mirror to your feet is similar to the triangle from the tower's top to the mirror to its base. Measuring the two ground distances and your own eye height gives the tower's height.

Proof

Let the observer's eye be at B, feet at A, the mirror at M, the tower base at C, and the tower top at D. By the law of reflection, ∠BMA = ∠DMC. Both triangles have a right angle at the ground, so △ABM ∼ △DCM. Hence DC / CM = AB / AM, giving the tower height DC = AB · CM / AM. Q.E.D.

Depth of a Well

To find the depth of a well by means of the sun.

OpticsProp. 21 Lay a straight bar across the mouth of the well. The sun casts the bar's shadow onto the well floor. By measuring the bar's position across the opening and the horizontal distance from the bar to where its shadow falls, you form a right triangle whose vertical leg is the well's depth. The sun angle is the same as in the shadow method (Prop. 19), so a stick above ground gives you that angle.

Proof

Let the well have width AB at the top and depth AD. Place a bar at E across the opening. The sun ray through E strikes the bottom at F. The horizontal offset BF (from the wall to the shadow) and the depth BD form a right triangle with the same sun angle as above ground. Hence BD / BF = h / s (from Prop. 19), giving the depth BD = h · BF / s. Q.E.D.

Length by Sighting Rod

To find the length of a wall from a distance by means of a rod.

OpticsProp. 22 Hold a rod of known length at arm's length and sight past both ends to the far wall. The rod and the wall, together with the two visual rays from the eye, form similar triangles. Knowing the rod length, the arm's length (eye-to-rod distance), and the distance to the wall, you compute the wall's length.

Proof

Let the eye be at E, the rod R₁R₂ at distance d, and the wall AB at distance D. The visual rays from E through R₁ and R₂ hit the wall at A and B. By similar triangles, AB / R₁R₂ = D / d, so AB = R₁R₂ · D / d. Q.E.D.

Ratio

A ratio is a sort of relation in respect of size between two magnitudes of the same kind.

ElementsV. Def. 3 A ratio captures the relative size of two magnitudes. We do not say what a ratio is in isolation—only that two magnitudes of the same kind (both lengths, both areas, etc.) stand in a ratio to one another. The ratio A : B tells us how A compares to B in size.

Euclid deliberately avoids reducing a ratio to a number. Two line segments have a ratio even when their lengths are incommensurable—that is, when no unit of length measures both exactly. This abstraction, due to Eudoxus, is what makes Book V applicable to all magnitudes, rational or irrational.

Discussion

Two magnitudes A and B must be of the same kind to form a ratio: we can compare a length to a length, but not a length to an area. Given two such magnitudes, we write the ratio as A : B. In the figure below, A and B are line segments, and the ratio label shows their relationship.

Proportions

Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order.

ElementsV. Def. 5 This is the famous Eudoxian definition of proportion. Four magnitudes A, B, C, D are in proportion—written A : B :: C : D—when for every pair of positive integers m and n:

• if mA > nB, then mC > nD;
• if mA = nB, then mC = nD;
• if mA < nB, then mC < nD.

This definition bypasses the problem of incommensurable magnitudes entirely. We never need to express the ratio as a fraction—we only test every possible multiple. It is the ancient equivalent of Dedekind cuts, predating them by over two millennia.

Magnitudes which have the same ratio are called proportional.

ElementsV. Def. 6 When the proportion A : B :: C : D holds, we say the four magnitudes are proportional.

Discussion

In the figure below we show two pairs of magnitudes and then test them with multiples m = 2 and n = 3. Because the same ordering holds for the multiples of each pair, the ratios are equal.

Duplicate Proportion

When three magnitudes are proportional, the ratio of the first to the third is said to be the duplicate ratio of the first to the second.

ElementsV. Def. 9 If A, B, C are in continued proportion—meaning A : B :: B : C—then the ratio A : C is called the duplicate of the ratio A : B. In modern terms, the duplicate ratio corresponds to squaring: if A : B = r, then A : C = r².

This concept is essential for the theory of similar figures in Book VI, where areas of similar polygons stand in the duplicate ratio of their corresponding sides.

Discussion

Below, A : B has ratio 3 : 2 (segments of length 180 and 120). Since B : C is also 3 : 2, C has length 80. The duplicate ratio A : C is 9 : 4.

Distributivity

If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth.

ElementsV.2 In modern notation: if nA and mA are multiples of A, and nB and mB are the corresponding multiples of B, then (n+m)A is the same multiple of A that (n+m)B is of B. That is, nA + mA = (n+m)A.

This is the distributive law for multiples of magnitudes. It may seem obvious, but Euclid proves it carefully because magnitudes are not assumed to be numbers. Every algebraic property must be established from the definitions.

Proof

Let A and B be magnitudes. Let nA consist of n copies of A, and let mA consist of m copies of A. Then nA + mA consists of n + m copies of A, so nA + mA = (n+m)A. The same argument applies to B, giving nB + mB = (n+m)B. Q.E.D.

Associativity of Multiplication

If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth.

ElementsV.3 In modern terms: m(nA) = (mn)A. If we first take n copies of A, then take m copies of that result, we get mn copies of A. This is the associativity (or compatibility) of scalar multiplication with magnitudes.

Proof

Let nA be n copies of A. Then m(nA) consists of m groups, each containing n copies of A. Altogether there are mn copies of A, so m(nA) = (mn)A. Q.E.D.

Transitivity of Proportions

Ratios which are the same with the same ratio are also the same with one another.

ElementsV.11 If A : B = C : D and C : D = E : F, then A : B = E : F. Proportion is a transitive relation. This is used constantly throughout the Elements whenever chains of equal ratios are linked together.

Proof

Take any positive integers m and n. Since A : B = C : D, if mA > nB then mC > nD. Since C : D = E : F, if mC > nD then mE > nF. Therefore if mA > nB then mE > nF. The same chain holds for equality and for the less-than case. By Definition V.5, A : B = E : F. Q.E.D.

Alternate Proportions

If four magnitudes be proportional, they will also be proportional alternately.

ElementsV.16 If A : B = C : D, then A : C = B : D. We may “swap the means” of a proportion. This is one of the most frequently used manipulation rules in the Elements, enabling us to shift between comparing like magnitudes and comparing corresponding magnitudes.

Proof

Take any positive integers m and n. Because A : B = C : D, we know mA compares to nB the same way mC compares to nD (by Def. V.5). But mA and mC are equimultiples of A and C, while nB and nD are equimultiples of B and D. So mA compares to nB the same way mC compares to nD—which is exactly the condition for A : C = B : D. Q.E.D.

Ratios ex aequali

If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali.

ElementsV.22 Given two sequences of magnitudes A, B, C and D, E, F with A : B = D : E and B : C = E : F, then A : C = D : F. This is the “chain rule” for proportions—intermediate terms cancel when ratios are composed ex aequali (“through equals”).

The result generalizes to any number of terms: given A₁, …, A_n and B₁, …, B_n with each consecutive pair in the same ratio, the first-to-last ratios are equal. Euclid uses this proposition extensively in Books VI and XII.

Proof

Take any positive integers m and n. Since A : B = D : E, by Def. V.5 if mA > nB then mD > nE. Since B : C = E : F, for any p, if nB > pC then nE > pF. Choose n so that the comparison of mA to pC passes through B: the chain mA ≥ nB ≥ pC forces mD ≥ nE ≥ pF. The same holds for equality and less-than. Therefore A : C = D : F. Q.E.D.

Similarity

Similar rectilineal figures are such as have their several angles equal, each to each, and the sides about the equal angles proportional.

ElementsVI. Def. 1 Two polygons are similar when they have exactly the same shape but may differ in size. This means two things must hold at once: every angle in one figure equals the corresponding angle in the other, and the sides surrounding each pair of equal angles are in the same ratio.

Similarity is the central concept of Book VI. Where Book I studied congruence—figures that are equal in every respect—similarity relaxes the requirement on size while keeping the requirement on shape. A small triangle can be similar to a large one; a photograph can be similar to the scene it depicts.

Discussion

Consider triangles ABC and DEF. They are similar when ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, and the sides about each pair of equal angles are proportional: AB : DE = AC : DF = BC : EF. The figure below shows a pair of similar triangles, first highlighting their equal angles and then their proportional sides.

Triangle Areas Proportional to Bases

Triangles and parallelograms which are under the same height are to one another as their bases.

ElementsVI.1 This fundamental proposition links area to length. When two triangles share the same height, their areas are in the same ratio as their bases. This result converts a two-dimensional comparison (area) into a one-dimensional one (base length), and it underpins nearly every area argument in Book VI.

Proof

Let triangles PQR and STU have equal heights. Place them so that their bases PQ and ST lie on the same line. The area of each triangle equals half the product of its base and the common height. Therefore Area(PQR) : Area(STU) = PQ : ST. Q.E.D.

Parallel Cuts Sides Proportionally

If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally.

ElementsVI.2 A line drawn inside a triangle parallel to one side creates a smaller triangle similar to the original. The key consequence is that the line divides both other sides in the same ratio. This fact is the workhorse of similar-triangle arguments throughout geometry.

Proof

In triangle ABC, let DE be drawn parallel to BC, with D on AB and E on AC. Triangles BDE and BDC share base BD and have equal heights (since DE ∥ BC). By §7.2, their areas are as their bases. A similar argument for the other pair shows AD : DB = AE : EC. Q.E.D.

Angle Bisector Theorem

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle.

ElementsVI.3 The angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides. This elegant result links the angular geometry of a triangle to its side lengths through proportion.

Proof

In triangle ABC, let AD bisect angle A, meeting BC at D. Draw CE parallel to DA to meet BA produced at E. Since AD ∥ EC, the alternate angles show that triangle ACE is isosceles with AC = AE. By §7.3, BD : DC = BA : AE = BA : AC. Therefore BD : DC = AB : AC. Q.E.D.

Similar Triangles

Equiangular triangles have their sides about the equal angles proportional.

ElementsVI.4 If two triangles have two angles equal (and therefore all three, since the angle sum is two right angles by §1.9), then their corresponding sides are proportional. This is the AA criterion for similarity: two angle equalities suffice to guarantee that the triangles are similar.

ElementsVI.5 provides the converse: if the sides are proportional, the triangles are equiangular. Together, VI.4 and VI.5 establish that equal angles and proportional sides are equivalent conditions for triangles.

Proof

Let triangles ABC and DEF have ∠A = ∠D and ∠B = ∠E. Then ∠C = ∠F as well. Place DEF so that D coincides with A and DE lies along AB. Since ∠D = ∠A, side DF lies along AC. Because ∠E = ∠B, line EF is parallel to BC by §1.8. Then by §7.3, AB : DE = AC : DF = BC : EF. Q.E.D.

Side-Angle-Side Similarity

If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

ElementsVI.6 The SAS similarity criterion mirrors the SAS congruence criterion (§1.2) but replaces “equal sides” with “proportional sides.” If one angle of a triangle equals one angle of another, and the sides enclosing those angles are in proportion, the triangles are similar.

Proof

Let ∠A = ∠D and AB : DE = AC : DF. On DE construct angle DEG equal to ∠B, making triangle DEG equiangular with ABC. By §7.5 their sides are proportional, so AB : DE = AC : DG. Combining with the hypothesis gives DF = DG, whence triangle DEF is congruent to triangle DEG and therefore equiangular with ABC. Q.E.D.

Divide Line Into Equal Parts

From a given straight line to cut off a prescribed part.

ElementsVI.9 This construction shows how to divide a line segment into any number of equal parts using only compass and straightedge. The method draws an auxiliary ray, steps off equal lengths along it, and then uses parallels to transfer the equal division back to the original segment.

Construction

Given segment AB. Draw any ray from A not collinear with AB. With a compass, mark off three equal segments AC₁, C₁C₂, C₂C₃ on the ray. Join C₃ to B. Through C₁ and C₂ draw lines parallel to C₃B. By §7.3, these parallels divide AB into three equal parts. Q.E.F.

Divide Line In Ratio

To cut a given uncut straight line similarly to a given cut straight line.

ElementsVI.10 Where §7.7 divides a segment into equal parts, this proposition divides it in any prescribed ratio. The technique is the same—auxiliary ray, stepped-off lengths, and a parallel—but the lengths stepped off correspond to the terms of the desired ratio rather than being equal.

Construction

Given segment AB and ratio 2 : 1. Draw a ray from A and mark three equal segments on it (two for the first term, one for the second). Join the last mark to B. Through the second mark draw a line parallel to this join. By §7.3, it meets AB at point D with AD : DB = 2 : 1. Q.E.F.

Mean Proportionals

To two given straight lines to find a mean proportional.

ElementsVI.13 Given two segments a and b, their mean proportional is a segment h such that a : h = h : b. Equivalently, h² = a · b. The construction uses a semicircle: place a and b end-to-end as a diameter, then draw the perpendicular from the junction to the semicircle.

ElementsVI.12 shows that the perpendicular from the right angle to the hypotenuse creates the same mean-proportional relationship, which is the theoretical foundation for this construction.

Construction

Place segments a and b end-to-end along a line, forming diameter a + b. Construct a semicircle on this diameter. At the junction of a and b, erect a perpendicular to meet the semicircle at H. The segment from the junction to H is the mean proportional h = √(a · b). Q.E.F.

Fourth Proportional

If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional.

ElementsVI.16 This proposition connects proportions with rectangles. If w : x = y : z, then the rectangle w × z has the same area as the rectangle x × y. This “cross-multiplication” of proportions is used repeatedly in the rest of Book VI.

Proof

Given segments w, x, y in proportion w : x = y : z, the fourth proportional z satisfies w · z = x · y. This is proved by constructing the two rectangles and showing they have equal area using the theory of proportions from §6.1. The converse follows by reversing the argument. Q.E.D.

Areas of Similar Triangles

Similar triangles are to one another in the duplicate ratio of the corresponding sides.

ElementsVI.19 When two triangles are similar with sides in ratio k, their areas are in ratio k². This is a special case of the general principle that areas scale as the square of linear dimensions, which was defined abstractly as duplicate proportion in §6.3.

Proof

Let triangles ABC and DEF be similar with AB : DE = k. Find a third proportional BG such that AB : DE = DE : BG. Then AB : BG is the duplicate ratio of AB : DE. Since the triangles are similar, AB : DE = BC : EF. By §7.2, Area(ABC) : Area(ABG) = BC : BG, and the result follows. Q.E.D.

Application of Areas

To describe a rectilineal figure similar to one given rectilineal figure and equal to another.

ElementsVI.25 This remarkable construction combines similarity and area. Given any polygon and any model shape, we can construct a figure that is similar to the model but has area equal to the given polygon. It draws on the mean proportional (§7.9) and area results from earlier sections.

Construction

Let P be the given polygon with area S, and let T be the model triangle. Convert P into a rectangle with the same area (by §1.14), then convert the model T into a rectangle with the same area and the same height. The widths of these rectangles define two segments; their mean proportional (§7.9) gives the scaling factor needed to construct a triangle similar to T with area S. Q.E.F.

Generalizing the Pythagorean Theorem

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

ElementsVI.31 The Pythagorean theorem (§1.15) states that the square on the hypotenuse equals the sum of the squares on the legs. But there is nothing special about squares—any similar figures will do. If we draw similar shapes on the three sides of a right triangle, the figure on the hypotenuse has area equal to the sum of the figures on the two legs.

Proof

Let triangle ABC have a right angle at A. Drop the altitude AH to the hypotenuse BC. This creates two sub-triangles ABH and ACH, each similar to the whole triangle ABC (by AA, since each shares an angle with ABC and has a right angle). By §7.11, areas of similar figures on corresponding sides are in the duplicate ratio. Since BH + HC = BC, the similar figures on AB and AC sum to the similar figure on BC. Q.E.D.

Commensurability

Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.

ElementsX. Def. 1 This definition is the gateway to everything that follows. Two line segments are commensurable if some smaller segment fits a whole number of times into each. For example, segments of length 10 and length 6 are commensurable because a segment of length 2 fits exactly 5 times into the first and 3 times into the second. Their ratio is therefore 5 : 3 — a ratio of whole numbers.

When no common unit exists — no matter how small — the magnitudes are incommensurable. The discovery that such pairs exist was one of the great shocks of ancient mathematics, overturning the Pythagorean belief that “all is number.”

Discussion

The concept is closely related to the Euclidean algorithm of Chapter 7: two segments are commensurable precisely when the process of repeated subtraction terminates. If it never terminates — if we keep finding remainders forever — the segments are incommensurable. This link between arithmetic process and geometric fact runs through the entire chapter.

Exhaustion

Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.

ElementsX.1 This is the method of exhaustion — perhaps the most far-reaching single proposition in all of Greek mathematics. It says that by repeatedly removing more than half of what remains, any magnitude can be made smaller than any given target, no matter how tiny. Archimedes would later use this principle to compute areas and volumes that we now handle with integral calculus.

The proof is a beautiful example of indirect reasoning. Suppose the remainder never falls below the target ε. Then at each step the magnitude decreases by more than half, so after n steps it is less than M / 2ⁿ. But 2ⁿ grows without bound, so eventually M / 2ⁿ < ε — a contradiction.

Proof

Let magnitude M and target ε be given with M > ε. Subtract from M a part greater than its half, leaving a remainder r₁ < M/2. Subtract from r₁ more than its half, leaving r₂ < M/4. Continue: after n steps the remainder rₙ < M / 2ⁿ. Since 2ⁿ can be made arbitrarily large, we eventually obtain rₙ < ε. Q.E.D.

Incommensurability

If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.

ElementsX.2 This proposition gives the definitive test for incommensurability: apply the Euclidean algorithm to two magnitudes. If the process of alternate subtraction never terminates — if there is always a remainder — then no common unit can exist.

The most famous application is the diagonal and side of a square. If the side has length 1, the diagonal has length √2. Subtracting the side from the diagonal leaves √2 − 1. Subtracting this remainder from the side leaves 2 − √2. The pattern continues without end, each remainder spawning a smaller copy of the same relationship. This is the geometric heart of the proof that √2 is irrational.

Proof

Let ABCD be a square with side s and diagonal d. Subtract s from d: the remainder is r₁ = d − s. Subtract r₁ from s: the remainder is r₂ = 2s − d. But r₂ and r₁ are related exactly as s and d were — they are the side and diagonal of a smaller square. The process therefore never terminates. By Proposition X.2, s and d are incommensurable. Q.E.D.

Square vs. Length Commensurability

Squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length.

ElementsX.9 This proposition links two seemingly different notions: whether two segments have a common measure (commensurability “in length”) and whether the squares built on them have a ratio expressible as a perfect-square fraction. If sides are in the ratio m : n, the squares are in the ratio m² : n² — a ratio of square numbers. Conversely, if the squares’ ratio is a ratio of square numbers, then the sides must be commensurable.

This result also has an important contrapositive: if two segments are incommensurable in length, then the ratio of their squares is not a ratio of square numbers. This is exactly the situation with the side and diagonal of a square, where the ratio of the squares is 1 : 2 and 2 is not a perfect square.

Proof

Let segments A and B be commensurable in length with A : B = m : n. The square on A has area proportional to m², and the square on B to n². Hence Sq(A) : Sq(B) = m² : n², a ratio of square numbers. Conversely, if Sq(A) : Sq(B) = p² : q², then A : B = p : q, so the sides are commensurable. Q.E.D.

Transitivity of Commensurability

Magnitudes commensurable with the same magnitude are commensurable with one another also.

ElementsX.12 This proposition establishes that commensurability is transitive: if A is commensurable with B, and B is commensurable with C, then A is commensurable with C. The proof proceeds through ratios: since A : B and B : C are both ratios of whole numbers, so is A : C.

Together with the reflexive and symmetric properties (which are immediate from the definition), transitivity makes commensurability an equivalence relation. The magnitudes of Book X are thereby partitioned into classes — those within each class can be measured by a common unit, while magnitudes in different classes cannot.

Proof

Let A be commensurable with B, so A : B = m : n for integers m, n. Let B be commensurable with C, so B : C = p : q. Then A : C = (A : B) · (B : C) = mp : nq, a ratio of integers. Hence A and C are commensurable. Q.E.D.

Difference of Medials Irrational

If from a medial area a medial area be subtracted, the side of the remaining area becomes one of two irrational straight lines, either a first apotome of a medial or a second apotome of a medial.

ElementsX.26 A medial area is one whose side is a mean proportional between two commensurable lines — in modern terms, an area like √6 whose square root is irrational. This proposition shows that when one medial area is subtracted from another, the result is again irrational. Irrationality, in a sense, is “closed under subtraction” within the medial class.

Euclid’s classification of irrationals is remarkably subtle. He identifies thirteen distinct species of irrational line, of which the medial is the simplest. The “first apotome of a medial” and “second apotome of a medial” are two further species that arise as remainders when medial areas are subtracted.

Proof

Let rectangle R₁ be a medial area and rectangle R₂ a smaller medial area. Both R₁ and R₂ have sides that are mean proportionals between commensurable lines. Their difference R₁ − R₂ is a rectangle whose side cannot be rational (otherwise R₁ and R₂ would share a common rational measure, contradicting their medial character). Hence the side of the remaining area is irrational. Q.E.D.

Two Squares Summing to a Square

To find two square numbers such that their sum is also square.

ElementsLemma X.29 This lemma, embedded in the proof of Proposition X.29, gives the geometric construction behind Pythagorean triples. Euclid shows that if you take any right triangle with legs a and b, the square on the hypotenuse c equals the sum of the squares on the legs: a² + b² = c². This is, of course, the Pythagorean theorem of Section 1.15, applied here to the arithmetic of square numbers.

The lemma works in reverse too: given any two squares whose sum is a square, we can always find a right triangle with those squares on its sides. This tight connection between geometry and arithmetic — between right angles and sums of squares — is one of the deepest themes in all of mathematics.

Proof

Construct a right triangle with legs of length a and b. By the Pythagorean theorem (I.47), the square on the hypotenuse c equals the sum of the squares on a and b. Hence a² + b² = c², and c² is a square number when a² and b² are square numbers. Q.E.D.

Sphere

When, the diameter of a semicircle remaining fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere.

ElementsXI. Def. 14 A sphere is the simplest solid — generated by rotating a semicircle around its diameter. Every point on the surface is equidistant from the centre. This definition by rotation connects plane geometry (the circle) to solid geometry in the most natural way possible.

Euclid’s definition is purely constructive: rather than listing properties, he tells you how to make a sphere. This approach — defining solids as swept-out plane figures — is the ancestor of every “solid of revolution” in modern calculus.

Discussion

The key insight is that a circle is a one-dimensional curve bounding a two-dimensional disk, while a sphere is a two-dimensional surface bounding a three-dimensional ball. The semicircle’s rotation sweeps out the full surface. Any plane through the centre cuts the sphere in a great circle equal to the generating circle; planes not through the centre produce smaller circles.

Regular Polyhedra

A cube is a solid figure contained by six equal squares. An octahedron is a solid figure contained by eight equal equilateral triangles. An icosahedron is a solid figure contained by twenty equal equilateral triangles. A dodecahedron is a solid figure contained by twelve equal equilateral pentagons.

ElementsXI. Def. 25–28 These four definitions, together with the tetrahedron (a pyramid of four equilateral triangles, defined earlier), complete the catalogue of the five regular polyhedra — also called the Platonic solids. A regular polyhedron has all faces congruent regular polygons, with the same number of faces meeting at every vertex.

That exactly five such solids exist is one of the most beautiful results in all of mathematics. Euclid proves this in the final proposition of the Elements (XIII.18), making it the climax of the entire work. The five solids are: the tetrahedron (4 triangular faces), the cube (6 square faces), the octahedron (8 triangular faces), the icosahedron (20 triangular faces), and the dodecahedron (12 pentagonal faces).

Discussion

Why only five? At each vertex, the face angles meeting there must sum to less than 360° (otherwise the faces would lie flat or overlap). For equilateral triangles (60° each), we can fit 3, 4, or 5 around a vertex — giving the tetrahedron, octahedron, and icosahedron. For squares (90°), only 3 fit — the cube. For regular pentagons (108°), only 3 fit — the dodecahedron. Hexagons (120°) already sum to 360° with just 3, so no regular solid is possible. This argument from Section 9.7 on solid angles makes the classification inevitable.

Intersection of Planes

If two planes cut one another, their common section is a straight line.

ElementsXI.3 When two flat surfaces meet in space, they share a perfectly straight line — and nothing more. This is so intuitive that it hardly seems to need proof, yet Euclid proves it rigorously: any point common to both planes lies on the line joining two such common points, because each plane, being flat, contains the whole of that line.

This result is foundational for everything that follows. It means we can always describe the meeting of two planes by a single line, reducing a three-dimensional situation to a one-dimensional one.

Proof

Let planes α and β meet, and let A and B be two points in their intersection. Both planes contain the line AB (since a plane contains the whole of any line whose endpoints lie in it). Suppose some point C lies in both planes but not on line AB. Then both planes contain the triangle ABC, so they coincide — contradicting our assumption that they are distinct. Therefore the intersection is exactly the line AB. Q.E.D.

Two Perpendiculars to a Plane Are Parallel

If two straight lines be set up at right angles to the same plane, the straight lines will be parallel.

ElementsXI.6 In plane geometry, two lines perpendicular to the same line are parallel. This proposition is the three-dimensional analogue: two lines perpendicular to the same plane must be parallel. It is the fundamental link between perpendicularity and parallelism in space.

The proof works by contradiction. If the two perpendicular lines were not parallel, they would determine a plane that intersects the given plane in a line. But then both lines would be perpendicular to that intersection line while lying in the same plane — which means they must be parallel in that plane, contradicting the assumption.

Proof

Let lines AB and CD both be perpendicular to plane α, with B and D in α. Join BD. Since AB ⊥ α, we have AB ⊥ BD. Similarly CD ⊥ BD. Two lines in the plane ABDC both perpendicular to BD at its endpoints are parallel (I.28). Hence AB ∥ CD. Q.E.D.

Lines Perpendicular to Planes

From a given elevated point to draw a straight line perpendicular to a given plane. — XI.11
To set up a straight line at right angles to a given plane from a given point in it. — XI.12

ElementsXI.11, XI.12 These companion propositions solve the two natural construction problems for perpendiculars in space: dropping a perpendicular from a point above a plane (XI.11), and erecting a perpendicular at a point in the plane (XI.12). Together they guarantee that through any point in space, exactly one line can be drawn perpendicular to a given plane.

The construction of XI.12 is particularly elegant: draw any line in the plane through the given point, erect a perpendicular to that line in the plane, then erect a perpendicular to that line at the given point. This final line is perpendicular to the entire plane, because it is perpendicular to two independent lines in it.

Proof

XI.11: Let P be a point above plane α. Draw any line DE in α and drop perpendicular PF to DE (I.12). At F in the plane, draw GH ⊥ DE. Drop perpendicular PK from P to GH. Then PK ⊥ α.

XI.12: Let A be a point in plane α. Draw line BC through A in the plane. At A, erect AD ⊥ BC in the plane. At A, erect AE ⊥ AD out of the plane. Since AE is perpendicular to two lines in α, it is perpendicular to α. Q.E.D.

Planes Perpendicular to the Same Line Are Parallel

Planes to which the same straight line is at right angles will be parallel.

ElementsXI.14 This is the spatial analogue of a familiar plane result: just as lines perpendicular to the same line are parallel (I.28), planes perpendicular to the same line are parallel. Two such planes cannot meet, because if they did their intersection line would be perpendicular to the given line in two different planes — impossible by XI.6.

Proof

Let line AB be perpendicular to both plane α and plane β. Suppose α and β meet. Their intersection is a line CD (by XI.3). At any point E on CD, line AB is perpendicular to CE (since AB ⊥ α) and also perpendicular to DE (since AB ⊥ β). But CE and DE are the same line, so AB is perpendicular to CD from two points — a contradiction. Hence α ∥ β. Q.E.D.

Solid Angles

To construct a solid angle out of three plane angles, two of which, taken together in any manner, are greater than the remaining one: thus the three angles must be less than four right angles.

ElementsXI.23 A solid angle is the three-dimensional analogue of a plane angle: where a plane angle is formed by two rays meeting at a point, a solid angle is formed by three or more plane faces meeting at a vertex. Think of the corner of a room, where three walls (or two walls and the ceiling) meet.

Euclid’s constraint is crucial: the plane angles at the vertex must sum to less than 360°. If they sum to exactly 360°, the faces lie flat in a plane and no solid angle forms. If they sum to more, the faces would overlap. This is the key to the classification of regular polyhedra in Section 9.2.

Proof

Let three plane angles α, β, γ be given with α + β + γ < 360°, and each pair summing to more than the third (the triangle inequality for angles). Construct triangles with these vertex angles and equal sides emanating from the vertex. The three outer edges form a triangle, and folding the three triangular faces upward around this base creates the solid angle. The constraint α + β + γ < 360° ensures the faces close upward rather than lying flat. Q.E.D.

Volumes of Parallelepipeds

Parallelepipedal solids which are on equal bases and of the same height are equal to one another.

ElementsXI.32 A parallelepiped is the three-dimensional analogue of a parallelogram: a solid with six faces, each a parallelogram, with opposite faces parallel and congruent. A rectangular box is a special case, but the general parallelepiped can be “sheared” in any direction.

This proposition is the spatial version of I.35 (parallelograms on the same base and between the same parallels are equal in area). Just as shearing a parallelogram preserves area, shearing a parallelepiped preserves volume. The base area and height alone determine the volume.

Proof

Let P₁ and P₂ be parallelepipeds on equal bases with the same height. Align their bases on the same plane. Since each face is a parallelogram, the solids can be decomposed into matching prisms that rearrange to show equality, just as parallelograms are shown equal by rearranging triangles. Hence Vol(P₁) = Vol(P₂). Q.E.D.

Volumes of Prisms

Equal prisms which have triangular bases have also their bases reciprocally proportional to their heights; and those prisms which have triangular bases and their bases reciprocally proportional to their heights are equal.

ElementsXI.39 A prism with a triangular base is exactly half a parallelepiped, split along a diagonal plane. This fundamental relationship reduces all prism-volume questions to parallelepiped-volume questions, which are handled by XI.32.

The proposition also gives the reciprocal relationship: two prisms are equal in volume precisely when their bases and heights are inversely proportional. If one prism has twice the base area, it needs half the height to match the other’s volume. This extends the area reciprocity of I.35 and I.41 into the third dimension.

Proof

Let T be a triangular prism. Complete it to a parallelepiped P by reflecting T across the diagonal plane. Then Vol(T) = ½ Vol(P). If two prisms T₁ and T₂ have equal volumes, then their completed parallelepipeds also have equal volumes. By XI.32, their bases and heights must satisfy the reciprocal proportion: Base(T₁) : Base(T₂) = Height(T₂) : Height(T₁). Q.E.D.

Circular Areas

Circles are to one another as the squares on their diameters.

ElementsXII.2 Double the diameter of a circle and its area quadruples — not doubles. This proposition makes that scaling law rigorous. It is the first result in the Elements to measure a curved figure, and Euclid’s proof introduces the method of exhaustion in its fully mature form: inscribe regular polygons of 4, 8, 16, … sides; each doubling removes more than half the remaining gap between polygon and circle (X.1); a double reductio then eliminates both “greater” and “less.”

The result underpins every later volume formula in this chapter. Once circles scale as the square of the diameter, cones and cylinders inherit the same base-scaling rule, and the sphere formula (XII.18) ultimately rests on this foundation.

Proof

Let circles C₁ and C₂ have diameters d₁ and d₂. Inscribe a square in C₁; its area is ½ d₁², which exceeds half the circle. Bisect each arc and join the new vertices to form a regular octagon; each new triangle exceeds half the circular segment it replaces. Continue doubling: at every stage the inscribed polygon eats more than half the remaining gap. By X.1, the residual circle-minus-polygon can be made smaller than any assigned magnitude.

Suppose C₁ : C₂ > d₁² : d₂². Then some region S smaller than C₂ satisfies C₁ : S = d₁² : d₂². Inscribe a polygon P₂ in C₂ with residual less than C₂ − S, so P₂ > S. Inscribe a similar polygon P₁ in C₁. Similar polygons scale as the squares of corresponding diameters, so P₁ : P₂ = d₁² : d₂² = C₁ : S. But P₁ < C₁ while P₂ > S — a contradiction. The symmetric argument eliminates “less than,” leaving C₁ : C₂ = d₁² : d₂². — Q.E.D.

Triangular Prism Division

Every prism which has a triangular base is divided into three pyramids equal to one another which have triangular bases.

ElementsXII.7 A triangular prism can be sliced into exactly three tetrahedra of equal volume. This elegant dissection is the key to proving that any pyramid is one-third of the prism sharing its base and height — the three-dimensional analogue of a triangle being half its enclosing parallelogram (I.41).

Proposition XII.6 establishes the preliminary lemma: a prism on a triangular base can be cut by two planes into three pyramids, and at least two of them are equal. Proposition XII.7 then shows all three are equal, yielding the celebrated “one-third” volume rule for pyramids. This result is the stepping stone to the cone formula in XII.10.

Proof

Let prism ABCDEF have triangular base ABC and top face DEF with AD, BE, CF as lateral edges. Cut with planes BDE and BEC. This produces three tetrahedra:

T₁ = ABDE,  T₂ = BCDE,  T₃ = BCEF.

Tetrahedra T₁ and T₂ share base triangle BDE and have equal heights from A and C respectively (since AC is parallel to the plane through BDE in the prism). So T₁ = T₂. Tetrahedra T₂ and T₃ share base triangle BCE and have equal heights from D and F respectively (since DF is parallel to the base plane). So T₂ = T₃. Therefore T₁ = T₂ = T₃, and each pyramid is one-third of the prism. — Q.E.D.

Cones

Any cone is a third part of the cylinder which has the same base with it and equal height.

ElementsXII.10 A cone fills exactly one-third of the cylinder that shares its circular base and height. This is the curved-surface analogue of the “pyramid = ⅓ prism” theorem (XII.7), and the proof follows the same strategy: inscribe polygonal pyramids inside the cone and polygonal prisms inside the cylinder, apply the one-third rule for each pair, then use the method of exhaustion to pass from polygons to circles.

The result was known to Democritus by physical intuition, but Eudoxus supplied the rigorous exhaustion proof that Euclid records here. Together with XII.2, it establishes that the volume of a cone is ⅓ π r² h — though Euclid, having no symbol for π, states everything as ratios.

Proof

Let cone V–ABCD and cylinder ABCD–EFGH share circular base ABCD and height h. Inscribe a square ABCD in the base circle. Erect the pyramid V–ABCD inside the cone and the prism ABCD–EFGH inside the cylinder. By XII.7, the pyramid is one-third of its prism.

Now double the sides: inscribe an octagon, then a 16-gon, and so on. Each inscribed pyramid is one-third of the corresponding inscribed prism. The inscribed prisms exhaust the cylinder and the inscribed pyramids exhaust the cone (each doubling removes more than half the remaining gap, so by X.1 the residual vanishes). Suppose the cone were not exactly one-third of the cylinder — then some magnitude S ≠ cone would satisfy cylinder = 3S. But the exhaustion squeezes S between inscribed pyramids and the cone, yielding a contradiction in both directions. Therefore cone = ⅓ cylinder. — Q.E.D.

The Height of a Cone/Cylinder

Cones and cylinders which are of the same height are to one another as their bases.

ElementsXII.11 Given two cylinders (or two cones) of equal height, their volumes are in the same ratio as their circular bases — which, by XII.2, means in the ratio of the squares of their diameters. The companion result (XII.14) establishes the dual: cylinders on equal bases are as their heights. Together they confirm the modern formula V = πr²h for a cylinder and V = ⅓πr²h for a cone, expressed in Euclid’s ratio language.

The proof combines XII.10 (cone = ⅓ cylinder) with the parallelepiped results of XI.32: inscribed prisms of equal height have volumes proportional to their bases; the exhaustion argument lifts this from prisms to cylinders; then the one-third rule transfers it to cones.

Proof

Let cylinders K₁ and K₂ have equal height h and circular bases with diameters d₁ and d₂. Inscribe regular n-gons in each base and erect prisms of height h. These prisms have equal heights, so their volumes are as their bases (XI.32). The inscribed polygonal bases are similar, hence as the squares of corresponding diameters. As n doubles, the prisms exhaust the cylinders. By the method of exhaustion, K₁ : K₂ = d₁² : d₂². Since each cone is one-third of its cylinder (XII.10), the same ratio holds for cones of equal height. — Q.E.D.

The Volume of a Sphere

Spheres are to one another as the cubes on their diameters.

ElementsXII.18 Double the diameter of a sphere and its volume increases eightfold. This is the ultimate proposition of Euclid’s measurement programme: circles scale as diameter-squared (XII.2), cylinders and cones add one linear factor for height (XII.11), and now spheres — curving in every direction — scale as the cube of the diameter. The result is the capstone of Book XII and one of the deepest applications of the method of exhaustion in the Elements.

Archimedes later proved the exact formula — a sphere is two-thirds of its circumscribing cylinder — but Euclid’s ratio theorem stands on its own: it tells us how spheres compare without ever computing an individual volume. The proof inscribes polyhedra whose volumes can be computed from pyramids, then uses the exhaustion technique to transfer the cube-scaling law from polyhedra to spheres.

Proof

Let spheres S₁ and S₂ have diameters d₁ and d₂. Inscribe a polyhedron in S₁ by stacking layers of pyramids whose apices lie at the centre. Each pyramid has its base on the polyhedral surface and height equal to the inradius. Doubling the number of faces at each stage, each doubling removes more than half the residual volume between polyhedron and sphere (by the same argument as XII.2).

Inscribe a similar polyhedron in S₂. Similar polyhedra have volumes in the ratio of the cubes of corresponding edges, and the edges scale with the diameters. So the inscribed polyhedra satisfy P₁ : P₂ = d₁³ : d₂³. Suppose S₁ : S₂ > d₁³ : d₂³. Then some S′ < S₂ satisfies S₁ : S′ = d₁³ : d₂³. Choose a polyhedron P₂ with residual less than S₂−S′, so P₂ > S′. Then P₁ : P₂ = d₁³ : d₂³ = S₁ : S′. But P₁ < S₁ and P₂ > S′ — contradiction. The symmetric argument eliminates “less than.” Therefore S₁ : S₂ = d₁³ : d₂³. — Q.E.D.

The Tetrahedron

If the side of the equilateral triangle inscribed in the circle be cut in extreme and mean ratio, the square on the side of the regular tetrahedron inscribed in the sphere is two-thirds of the square on the diameter of the sphere.

ElementsXIII.9 The tetrahedron is the simplest Platonic solid — four equilateral triangle faces, four vertices, six edges. Euclid inscribes it in a sphere by finding the centre, erecting a perpendicular from the centre of an equilateral triangle base to locate the apex. The edge-to-diameter ratio, e² = (⅔)d², fixes the tetrahedron’s size uniquely within the sphere.

The construction begins with the sphere’s diameter. Find a point dividing the diameter so that the upper segment is twice the lower. The circle at that height becomes the circumcircle of the triangular base; the top of the diameter is the apex. All four faces are equilateral because the construction forces all six edges equal.

Proof

Let the sphere have diameter d and centre O. Take diameter AB. Find point C on AB such that AC = 2·CB. The plane through C perpendicular to AB cuts the sphere in a circle; inscribe equilateral triangle DEF in that circle. Then ADEF is the tetrahedron. Since AC = 2·CB and AB = d, we have CB = d/3 and AC = 2d/3. The radius of the base circle r satisfies r² = CB·CA = (d/3)(2d/3) = 2d²/9. The edge e of the inscribed equilateral triangle is e = r√3, so e² = 3r² = 3·(2d²/9) = 2d²/3. Each lateral edge AD also equals e (verify by Pythagorean theorem on triangle ACD: AD² = AC² + CD² = (2d/3)² + (2d²/9) = 4d²/9 + 2d²/9 = 6d²/9 = 2d²/3 = e²). So all six edges are equal. — Q.E.D.

The Octahedron

If the square on the side of the regular octahedron inscribed in a sphere be set out, it is half the square on the diameter of the sphere.

ElementsXIII.10 The octahedron has eight equilateral triangle faces, six vertices, and twelve edges. Euclid constructs it by taking three mutually perpendicular diameters of the sphere and connecting their six endpoints. The edge-to-diameter ratio is e² = d²/2, equivalently e = d/√2.

This is the most symmetrical construction in Book XIII: the octahedron’s axes align with the sphere’s diameters, making it the dual of the cube (vertices of one correspond to face-centres of the other). Euclid proves regularity by showing every edge connects endpoints of two perpendicular diameters, hence all edges are equal.

Proof

Let the sphere have diameter d. Take two perpendicular diameters meeting at centre O, giving four endpoints A, B, C, D forming a square in a great circle. Erect a third diameter perpendicular to this plane through O, with endpoints E (top), F (bottom). Connect every pair of adjacent vertices. Each edge (e.g. AE) is a hypotenuse of a right triangle with legs d/2 and d/2, so e² = (d/2)² + (d/2)² = d²/2. All 12 edges are identical by symmetry. The eight faces (e.g. ABE, BCE, CDE, DAE, ABF, BCF, CDF, DAF) are equilateral. — Q.E.D.

The Cube

If the side of the cube inscribed in a sphere be set out, the square on it is one-third of the square on the diameter of the sphere.

ElementsXIII.11 The cube has six square faces, eight vertices, and twelve edges. Inscribed in a sphere, its space diagonal equals the sphere’s diameter. Since the space diagonal of a cube with edge e is e√3, we get e√3 = d, hence e² = d²/3.

Euclid’s construction starts from the sphere’s diameter, derives the edge length, builds a square base, and erects the cube. The cube is the most intuitive Platonic solid — its faces and right angles make it the three-dimensional analogue of the square.

Proof

Let the sphere have diameter d. The space diagonal of a cube with edge e satisfies (e√3)² = d², so e² = d²/3. Construct a square ABCD with side e. Erect perpendiculars of height e at each vertex to get the top face EFGH. Every vertex lies on the sphere: the distance from centre O to any vertex equals half the space diagonal = d/2. All faces are squares with side e. — Q.E.D.

The Icosahedron

To construct a regular icosahedron and comprehend it in a sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.

ElementsXIII.13 The icosahedron has twenty equilateral triangle faces, twelve vertices, and thirty edges. It is the most complex Platonic solid to construct and the one most deeply entangled with the golden ratio φ = (1+√5)/2. Two parallel regular pentagons (top and bottom), rotated 36° relative to each other, plus a top and bottom apex, give the twelve vertices.

The construction is the longest in the Elements. Euclid uses the golden ratio (from II.11) to position two rings of five vertices each on parallel planes, then caps them with single vertices at the poles. The edge is classified as a “minor” irrational — one of the exotic irrationals catalogued in Book X.

Discussion

The sphere’s diameter d determines the edge through e² = d²(10 − 2√5)/20. Start with the diameter. Cut it in golden ratio to find the heights of the two pentagonal rings. Inscribe a regular pentagon in the circle at each height, rotating the upper ring by 36°. Connect adjacent vertices of the two rings with edges, forming a band of 10 triangles. Cap with apex vertices at the poles, connecting each to the five nearest ring vertices. The result: 12 vertices, 30 edges, 20 equilateral triangle faces.

The Dodecahedron

To construct a regular dodecahedron and comprehend it in a sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome.

ElementsXIII.14 The dodecahedron has twelve regular pentagon faces, twenty vertices, and thirty edges. Euclid constructs it by starting with a cube and erecting pentagonal “roofs” (tent-like structures) on each face. The golden ratio governs the roof height: each roof peak extends by φ times half the cube edge.

The edge is classified as an “apotome” — another Book X irrational. The dodecahedron is the dual of the icosahedron: vertices of one correspond to face-centres of the other. Together with the icosahedron, it demonstrates how deeply the golden ratio penetrates three-dimensional geometry.

Discussion

Begin with a cube of edge a inscribed in the sphere. On each face, erect a “roof” — a ridge line of length a·φ centred on the face, perpendicular to one pair of edges. The roof extends above and below the face by a·(φ−1)/2. The 8 cube vertices plus 12 roof-ridge endpoints give the 20 vertices of the dodecahedron. Each original cube face becomes hidden inside; the visible surface consists of 12 regular pentagons. The edge of the dodecahedron is e = a/φ.

The Tetrahedron is Regular

To set out the sides of the five figures and compare them with one another. — The side of the pyramid is the power of two-thirds of the diameter of the sphere.

ElementsXIII.15 Having constructed the tetrahedron in XIII.9, Euclid now verifies it is genuinely regular: all four faces are equilateral triangles, and all four solid angles are congruent. The proof retraces the construction, confirming that all six edges are equal (as shown in XIII.9) and that each vertex is surrounded by exactly three equilateral triangles meeting at equal angles.

The solid angle at each vertex subtends 3 × 60° = 180° of face angle — the smallest total among all Platonic solids. This large “angular surplus” (360° − 180° = 180°) corresponds to the tetrahedron’s sharp, pointed shape.

Proof

From XIII.9, all six edges equal e with e² = 2d²/3. Each face is a triangle with three equal sides, hence equilateral. At each vertex, exactly three faces meet. By the symmetry of the construction (each vertex is related to the others by rotation), all four solid angles are congruent. The tetrahedron is therefore regular: 4 equilateral triangle faces, 4 vertices, 6 edges, solid angle sum 3 × 60° = 180° at each vertex. — Q.E.D.

The Cube, Octahedron, and Dodecahedron are Regular

To set out the sides of the five figures and to compare them with one another. — The side of the cube is the power of one-third of the diameter; the side of the octahedron, the power of one-half; the side of the dodecahedron, the apotome.

ElementsXIII.16 Euclid groups three regularity proofs into one proposition. For the cube: all faces are squares and all solid angles are right-angled corners (3 squares meeting at each vertex). For the octahedron: all faces are equilateral triangles and each vertex has four triangles meeting at equal angles. For the dodecahedron: all faces are regular pentagons and each vertex has three pentagons meeting.

These three solids have an elegant duality: the cube and octahedron are duals of each other (swapping vertices for face-centres), and the dodecahedron is dual to the icosahedron. The edge-to-diameter ratios — e² = d²/3 (cube), e² = d²/2 (octahedron), e = d/(φ√3) (dodecahedron) — establish a clean hierarchy of sizes within the same sphere.

Discussion

Each solid’s regularity follows from its construction. The cube (XIII.11): all edges e = d/√3 by construction, 6 square faces, 3 squares at each vertex with right angles. The octahedron (XIII.10): all edges e = d/√2, 8 equilateral triangle faces, 4 triangles at each vertex. The dodecahedron (XIII.14): all edges related through the golden ratio, 12 regular pentagon faces, 3 pentagons at each vertex. In each case, the construction forces all edges equal and all solid angles congruent.

The Icosahedron is Regular

To construct a dodecahedron and comprehend it in a sphere.

ElementsXIII.17 The icosahedron’s regularity proof is the most intricate in the Elements. With 20 faces, 30 edges, and 12 vertices, there are many equalities to verify. Euclid must show: (a) all 30 edges are equal, (b) all 20 faces are equilateral triangles, and (c) all 12 solid angles are congruent, each surrounded by exactly 5 triangles.

The proof relies heavily on the golden ratio properties established in XIII.8–12. The key insight is that the vertices form two parallel regular pentagons plus two poles, and the golden ratio ensures the inter-ring distances equal the intra-ring (edge) distances. The 5 triangles at each vertex sum to 5 × 60° = 300° of face angle, leaving an angular deficit of 60°.

Proof

From XIII.13, the 12 vertices lie on the sphere: 1 top pole, 5 upper ring, 5 lower ring (rotated 36°), 1 bottom pole. Within each ring, adjacent vertices are separated by the pentagon edge. Between rings, each upper vertex connects to two lower vertices at distance equal to the edge (verified using the golden ratio: the diagonal-to-side ratio of the pentagon is φ, and the inter-ring height is chosen so that the slant distance equals the edge). Each pole connects to all 5 nearest ring vertices at the same edge distance. Total: 30 equal edges, 20 equilateral faces, 5 triangles at each vertex. — Q.E.D.

There are Exactly Five Regular Solids

I say that, besides the aforesaid five figures, no other figure can be constructed which is contained by equilateral and equiangular figures equal to one another.

ElementsXIII.18 The grand finale of the Elements: no sixth Platonic solid exists. The proof is a beautiful combinatorial argument. At each vertex of a regular solid, several copies of the same regular polygon must meet. The total face angle at a vertex must be less than 360° (otherwise the faces would lie flat or overlap). This constraint, combined with the limited menu of regular polygons, leaves exactly five possibilities.

This is the first known “impossibility” proof in solid geometry. It transforms a seemingly open-ended question (“how many regular solids could there be?”) into a finite enumeration. The argument’s elegance has inspired mathematicians for over two millennia.

Proof

At each vertex of a regular solid, n copies of a regular p-gon meet. The interior angle of a regular p-gon is (p−2)·180°/p. The constraint is n × (p−2)·180°/p < 360°, i.e. n(p−2) < 2p, with n ≥ 3 and p ≥ 3. Enumerate:

• p = 3 (equilateral triangle, angle 60°): n can be 3 (tetrahedron), 4 (octahedron), or 5 (icosahedron). n = 6 gives 360° — flat, impossible.
• p = 4 (square, angle 90°): n can be 3 (cube). n = 4 gives 360° — flat.
• p = 5 (pentagon, angle 108°): n can be 3 (dodecahedron). n = 4 gives 432° — impossible.
• p = 6 (hexagon, angle 120°): n = 3 already gives 360° — flat. Impossible.
• p ≥ 7: even n = 3 exceeds 360°. Impossible.

Exactly five combinations work: {3,3}, {3,4}, {3,5}, {4,3}, {5,3}. Each has been constructed. No sixth regular solid exists. — Q.E.D.