Chapter 1

Euclid’s Elements, Book I (constructions) 2 Euclid’s Elements, Book I (constructions)

1.1 The use of ruler and compass

Euclid’s Elements can be read as a book on how to construct certain geometric figures efficiently and accurately using ruler and compass, and ascertaining the validity. The first three postulates before Book I are on the basic use of the ruler and the compass.

Postulate 1. To draw a straight line from any point to any point. With a ruler (straightedge) one connects two given points A and B to form the line (segment) AB, and there is only one such line. This uniqueness is assumed, for example, in the proof of I.4.

Postulate 2. To produce a finite straight line continuously in a straight line. Given two points A and B, with the use of a ruler one can construct a point C so that the line (segment) AC contains the point B. The first two postulates can be combined into a single one: through two distinct points there is a unique straight line. 1.1 The use of ruler and compass 3

Postulate 3. To describe a circle with any center and distance. This distance is given by a finite line (segment) from the center A to another point B. With the use of a collapsible compass, one constructs a circle with given center A to pass through B. We denote this circle by C(A, B). Euclid I.2 shows how to construct a circle with a given center and radius equal to a given line (seg- ment). 4 Euclid’s Elements, Book I (constructions)

Definition (I.20). Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.

Remark. Euclid seems to take isosceles and scalene in the exclusive sense. But it is more convenient to take these in the inclusive sense. An isosceles triangle is one with two equal sides, so that an equilateral triangle is also isosceles. 1.1 The use of ruler and compass 5

Euclid (I.1). On a given finite straight line to construct an equilateral triangle.

C

D E A B

Given: Points A and B. To construct: Equilateral triangle ABC. Construction: Construct the circles C(A, B) and C(B,A) to intersect at a point C. ABC is an equilateral triangle. 6 Euclid’s Elements, Book I (constructions)

The second proposition is on the use of a collapsible compass to transfer a seg- ment to a given endpoint.

Euclid (I.2). To place a straight line equal to a given straight line with one end at a given point.

D A

C B L

G

Given: Point A and line BC. To construct: Line AL equal to BC. Construction: (1) An equilateral triangle ABD, [I.1] (2) the circle C(B,C). (3) Extend DB to intersect the circle at G. (4) Construct the circle C(D, G) and (5) extend DA to intersect this circle at L. AL = DL − DA = DG − DB = BG = BC. Therefore the circle C(A, BC) can be constructed using a collapsible compass. 1.1 The use of ruler and compass 7

Euclid (I.3). Given two unequal straight lines, to cut off from the greater a straight line equal to the less. 8 Euclid’s Elements, Book I (constructions)

Euclid (I.9). To bisect a given rectilineal angle.

A

D E

F BC Given: Angle BAC. To construct: Line AF bisecting angle ABC. Construction: (1) Choose an arbitrary point D on AB. (2) Construct E on AC such that AD = AE. [I.3] (3) Construct an equilateral triangle DEF on DE (so that F and A are on opposite sides of DE). [I.1] The line AF is the bisector of the angle BAC. 1.1 The use of ruler and compass 9

Euclid (I.10). To bisect a given finite straight line.

C

A D B Given: Line segment AB. To construct: The midpoint of AB. Construction: (1) An equilateral triangle ABC, [I.1] (2) the bisector of angle ACB [I.9] (3) to meet AB at D. AD = DB. 10 Euclid’s Elements, Book I (constructions)

1.2 Perpendicular lines

Definition (I.10). When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.

Postulate 4. That all right angles are equal to each other.

Euclid (I.11). To draw a straight line at right angles to a given straight line from a given point on it.

F

AB D C E Given: A straight line AB and a point C on it. To construct: A line CF perpendicular to AB. Construction: (1) Take an arbitrary point D on AC, and construct E on CB such that DC = CE. (2) Construct an equilateral triangle DEF. [I.1] The line CF is perpendicular to AB. Proof : In triangles DCF and ECF, DC = EC, by construction CF = CF, DF = EF sides of equilateral triangle DCF ≡ ECF SSS ∠DCF = ∠ECF CF ⊥ AB. 1.2 Perpendicular lines 11

Euclid (I.12). To draw a straight line perpendicular to a given infinite straight line from a given point not on it.

C

AB G H E D

Given: A straight line AB and a point C not on the line. To construct: A line CH perpendicular to AB. Construction: (1) Take a point D on the other side of the line and construct the circle C(C, D) to meet the line at E and G. Bisect the segment EG at H [I.10] CH is perpendicular to AB. Proof : Triangles ECH and GCH are congruent. [SSS] ∠ECH = ∠GCH. CH ⊥ AB. 12 Euclid’s Elements, Book I (constructions)

1.3 Isosceles triangles

Euclid (I.5). In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.

A

B C

F G

DE

Given: Triangle ABC with AB = AC. To prove: ∠ABC = ∠ACB. Construction: Extend AB and AC to D and E respectively. Choose an arbitrary point F on BD, and construct G on CE such that CG = BF. [I.3] Proof : In triangles CAF and BAG, CA = BA, isosceles triangle ∠CAF = ∠BAG, AF = AB + BF = AC + CG = AG. ∴ BAG ≡CAF, SAS CF = BG, and ∠BFC = ∠AF C = ∠AGB = ∠CGB. In triangles BFC and CGB, BC = CB, ∠BFC = ∠CGB, CF = BG, proved above ∴ BFC ≡ CGB SAS ∠CBF = ∠BCG ∠ABC = ∠ACB. Q.E.D.

Q.E.D.: quod erat demonstrandum, which was to be demonstrated. 1.3 Isosceles triangles 13

This proposition is often called the “bridge of asses” (pons asinorum). According to D. E. Smith (The Teaching of Geometry, 1911, p.174), “[i]t is usually stated that it came from the fact that fools could not cross this bridge, and it is a fact that in the Middle Ages this was often the limit of the student’s progress in geometry. It has however been suggested that the name came from Euclid’s figure, which resembles the simplest type of a wooden truss bridge”. 14 Euclid’s Elements, Book I (constructions)

According to Proclus, Pappus (3rd century A.D.) proved the proposition in a very simple way by considering an isosceles triangle ABC as two triangles, ABC and ACB. From AB = AC, ∠BAC = ∠CAB, AC = AB, it follows from I.4 that ABC ≡ACB, and so ∠ABC = ∠ACB. 1.3 Isosceles triangles 15

Here is another apparently easy proof of I.5: let the bisector of angle BAC meet BC at D. Then ABD ≡ACD by I.4 again. From this we conclude that ∠ABD = ∠ACD, equivalently, ∠ABC = ∠ACB. However, the existence of an angle bisector is justified only in I.9. Euclid is a true constructivist. He does not consider squares before I.46. 16 Euclid’s Elements, Book I (constructions)

Euclid (I.6). If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

Given: Triangle ABC with ∠ABC = ∠ACB. To prove: AB = AC. Proof by contradiction: (1) Suppose AB is greater than AC. Choose a point D on AB such that DB = AC. [I.3] In triangles DBC and ACB, DB = AC, ∠DBC = ∠ACB, BC = CB. ∴ DBC ≡ ACB. SAS This is contradiction since triangle DBC is part of triangle ABC. (2) Suppose AB is less than AC. A similar reasoning also leads to a contradiction. (3) Therefore, AB = AC. Q.E.D. 1.4 Tests for congruence of triangles I.4,8,26 17

1.4 Tests for congruence of triangles I.4,8,26

Euclid did not use the term congruence of triangles. When he says two triangles are equal, he means they are equal in area.

Euclid (I.4). [SAS] If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remain- ing angles equal the remaining angles respectively, namely those which the equal sides subtend.

It is convenient to identify congruent triangles with corresponding vertices and sides. Thus, two triangles ABC and XY Z are congruent if the corresponding sides are equal, namely, BC = YZ, CA= ZX, XY = AB, and the corresponding angles are equal, namely, ∠BAC = ∠YXZ, ∠CBA = ∠ZY X, ∠ACB = ∠XZY.

Euclid I.4 is the first of four valid tests for congruence of triangles. We refer to it as the SAS test. Book I contains two more tests. 18 Euclid’s Elements, Book I (constructions)

Euclid (I.8). [SSS] If two triangles have the two sides equal to two sides respec- tively, and have also the base equal to the base, they will have the angles equal which are contained by the equal straight lines.

Euclid (I.26). [ASA, AAS] If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the re- maining angle. 1.4 Tests for congruence of triangles I.4,8,26 19

The following proposition, though not in Euclid’s Elements, completes the list of congruence tests.

Proposition (RHS). Two right angled triangles with equal hypotenuses and one pair of equal sides are congruent.

This is, of course, is a corollary of I.48. It can, however, be established without invoking this. Let ABC and ABC be two triangles in which the angles ACB and ACB are right angles, AB = AB and AB = AB. On the side of AC opposite to B, construct a triangle ADC congruent to ABC (with AD = AB and CD = CB). Since the angles ACB and ACD are both right angles, B, C, D are a line. Since AB = AD, by I.5, ∠ADC = ∠ABC. Therefore, triangles ABC and ADC are congruent (I.4); so are ABC and ABC. 20 Euclid’s Elements, Book I (constructions)

1.5 Two constructions

Euclid (I.22). Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.

Euclid (I.23). [Copying an angle along a line] On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectlineal angle.

Given: A line AB and an angle DCE. To construct: BAF equal to angle DCE. Construction: Choose arbitrary points D, E on the sides of the angle C. Construct triangle AF G such that CD = AF , CE = AG and DE = FG. [I.22] 1.6 Parallel lines 21

1.6 Parallel lines

Definition (I.23). Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.

Postulate 5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

This famous fifth postulate has a long history, eventually leading to the dis- covery of noneuclidean geometry in the 19th century. Here, we simply remark that it provides an explicit criterion for the intersection of finite lines upon extension. Heath [I.201] pointed out that while Postulate 4 can be proved using other axioms, “[i]t was essential from Euclid’s point of view that it should come before Post. 5, since the condition in the latter that a certain pair of angles are together less than two right angles would be useless unless it were first made clear that right angles are angles of determinate and invariable magnitude”. 22 Euclid’s Elements, Book I (constructions)

Euclid (I.27). If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. (I.28). If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. (I.29). A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. (I.30). Straight lines parallel to the same straight line are also parallel to one an- other. 1.6 Parallel lines 23

Euclid (I.31). Through a given point to draw a straight line parallel to a given straight line.

Given: A point A and a line BC not containing it. To construct: A line through A parallel to BC. Construction: (1) Take a point D on BC. (2) Copy angle ADC into angle DAE with E and C on opposite sides of AD. The line EA is parallel to BC. 24 Euclid’s Elements, Book I (constructions)

1.7 Parallelograms

Euclid (I.33). The straight lines joining equal and parallel straight lines [at the extremities which are] in the same directions are themselves equal and parallel.

Euclid (I.34). In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

A B

C D 1.7 Parallelograms 25

Euclid (I.35). Parallelograms which are on the same base and in the same parallels are equal to one another.

A D E F

B C 26 Euclid’s Elements, Book I (constructions)

Euclid (I.36). Parallelograms which are on equal bases and in the same parallels are equal to one another. (I.37). Triangles which are on the same base and in the same parallels are equal to one another. (I.38). Triangles which are on equal bases and in the same parallels are equal to one another. (I.39). Equal triangles which are on the same base and on the same side are also in the same parallels. (I.40). Equal triangles which are on equal bases and on the same side are also in the same parallels. (I.41). If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle. 1.7 Parallelograms 27

Euclid (I.42). To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

A F G

D B E C 28 Euclid’s Elements, Book I (constructions)

Euclid (I.43). In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

A H D

K E F

B G C 1.7 Parallelograms 29

Euclid (I.44). To a given straight line to “apply”, in a given rectilineal angle, a parallelogram equal to a given triangle.

F E K

C B D G M

H A L 30 Euclid’s Elements, Book I (constructions)

Euclid (I.45). To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

F G L D

C A

E

B K H M 1.8 The Pythagorean theorem 31

1.8 The Pythagorean theorem

Definition (I.22). Of quadrilateral figures, a square is that which is both equilat- eral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.

Euclid (I.46). To describe a square on a given straight line.

Given: A line AB. To construct: A square ABC. Construction: (1) Construct the circle C(A, B). (2) Construct the perpendicular to AB at A to meet the circle at D. (3) Construct the circles C(B,A) and C(D, A) to meet at C. ABCD is a square. 32 Euclid’s Elements, Book I (constructions)

Euclid (I.47). In right angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

K K

G G

H H A A

F F

C C B B

D L E D L E 1.8 The Pythagorean theorem 33

Book I ends with the converse of I.47.

Euclid (I.48). If in a triangle the square on one of the sides be equal to the squares on the remaining sides of the triangle, the angle contained by the remaining sides of the triangle is right.

Given: Triangle ABC with BC2 + CA2 = AB2. To prove: ∠ACB =1rt. ∠. Construction: Construct triangle ABC with ∠ACB =1rt. ∠ and AB = AB, AC = AC. Proof. AB2 = BC2 + AC2 = BC2 + AC2 = AB2. ∴ AB = AB. In triangles ABC and ABC, BC = BC, AC = AC, AB = AB, ∴ ABC ≡ABC [SSS] and ∠ACB = ∠ACB =1rt. ∠. Q.E.D.

Because of its usefulness, we reformulate the key step of Euclid’s proof of I.47 as follows. 34 Euclid’s Elements, Book I (constructions)

Proposition. Let the perpendicular from the right angle vertex C of a right triangle ABC intersect the hypotenuse AB at X. Then

AC2 = AX · AB, BC2 = BX · BA.

C

A X B 1.8 The Pythagorean theorem 35

Euclid (II.14). To construct a square equal to a given rectilineal figure.

H

B G E F

C D

Proof. (1) A rectilineal figure can be made equal to a rectangle [I.45]. (2) Conversion of a rectangle into a square:

H H

B B G E F G E F

C D C D 36 Euclid’s Elements, Book I (constructions)

1.9 Appendix: Tests of congruence of triangles

Construction of a triangle with three given elements A triangle has six elements: three sides and three angles. Consider the construction of a triangle given three of its six elements. The triangle is unique (up to size and shape) if the given data are in one of the following patterns. (1) SSS Given three lengths a, b, c, we construct a segment BC with length a, and the two circles B(c) and C(b). These two circles intersect at two points if (and only if) b + c>a[triangle inequality]. There are two possible positions of A. The resulting two triangles are congruent. (2) SAS Given b, c, and angle A<180◦, the existence and uniqueness of trian- gle ABC is clear. (3) ASA or AAS These two patterns are equivalent since knowing two of the angles of a triangle, we easily determine the third (their sum being 180◦). Given B, C and a, there is clearly a unique triangle provided B + C<180◦. (4) RHS Given a, c and C =90◦.

Congruence tests These data patterns also provide the valid tests of congruence of triangles. Two triangles ABC and XY Z are congruent if their corresponding elements are equal. Two triangles are congruent if they have three pairs of equal elements in one the five patterns above. The five valid tests of congruence of triangles are as follows. (1) SSS: ABC ≡XY Z if

AB = XY, BC = YZ, CA= ZX.

X A Z

B C Y

(2) SAS: ABC ≡XY Z if

AB = XY, ∠ABC = ∠XY Z, BC = YZ.

X A Z

B C Y 1.9 Appendix: Tests of congruence of triangles 37 X A Z

B C Y

(3) ASA: ABC ≡XY Z if ∠BAC = ∠YXZ, AB= XY, ∠ABC = ∠XY Z.

(4) AAS. We have noted that this is the same as ASA: ABC ≡XY Z if ∠BAC = ∠YXZ, ∠ABC = ∠XY Z, BC = YZ, .

X A Z

B C Y

(5) RHS. The ASS is not a valid test of congruence. Here is an example. The two triangles ABC and XY Z are not congruent even though ∠BAC = ∠YXZ, AB= XY, BC = YZ.

B Y

A C X Z However, if the equal angles are right angles, then the third pair of sides are equal: AC2 = BC2 − AB2 = YZ2 − XY 2 = XZ2, and AC = XZ. The two triangles are congruent by the SSS test. Without repeating these details, we shall simply refer to this as the RHS test. ABC ≡XY Z if ∠BAC = ∠YXZ =90◦,BC= YZ, AB= XY.

B Y

A C Z X