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.