The Elements

A series of four geometric diagrams are arranged vertically in the left margin.
1. Two side-by-side triangles. The left triangle has vertices labeled a, b, c, and a dot in the center. The right triangle has vertices labeled d, e, f.
2. A diagram showing two triangles sharing a vertex, with intersecting auxiliary lines. Vertices are labeled a, b, d, e, f, g.
3. A rhombus or kite-shaped figure formed by two triangles on a common base. Vertices are labeled a, b, c, d.
4. An isosceles triangle labeled a, b, c with a perpendicular line dropped from vertex c to the base at point d.

Similarly, they are equal by the same [proposition]; and because angle b.d.c. is greater than angle a.d.c., it follows that angle b.c.d. is greater than angle a.c.d.—a part [being greater than] the whole, which is impossible. This refers to the common mathematical axiom that the whole is always greater than any of its parts. If, however, point d falls outside triangle a.b.c. so that the lines do not intersect, I shall draw line d.c. and extend b.d., a.d., and b.c. below the base to points f and e. And since lines a.d. and a.c. are equal, angles a.c.d. and a.d.c. will be equal by the fifth [proposition]. Likewise, since b.c. and b.d. are equal, the angles below the base—which are c.d.f. and d.c.e.—will be equal by the second part of the same [proposition]. Therefore, because angle e.c.d. is smaller than angle a.c.d., it follows that angle f.d.c. is smaller than angle a.d.c., which is impossible. In the same way, the opponent A rhetorical term for someone proposing a counter-argument or an impossible geometric configuration. is led to an absurdity if point d falls inside triangle a.b.c., and so on.

Of any two triangles in which two sides of one are equal to two sides of the other, and the base of one is equal to the base of the other, it is necessary that the two angles contained by the equal sides be equal.
¶ Let there be two triangles, a.b.c. and d.e.f., and let a.c. be equal to d.f., b.c. equal to e.f., and a.b. equal to d.e. I say that angle c is equal to angle f, angle a to angle d, and angle b to angle e. I will superimpose base a.b. onto base d.e.; since they are equal, neither will exceed the other, according to the second-to-last common notion. original: "conceptionem"; referring to Euclid’s Common Notions, specifically that things which coincide with one another are equal. Therefore, either point c will fall upon point f, or it will not. If it does, then because angle c will be superimposed on angle f and neither exceeds the other, they are equal by the converse of the aforementioned notion. Similarly, argue that the remaining angles are equal. If, however, point c does not fall on f, but upon any other point, let it be point g. Because e.g. is equal to b.c. (indeed, it is the same), and likewise d.g. is equal to a.c., then e.g. will be equal to e.f. and d.g. will be equal to d.f., which is impossible by the preceding [proposition].

To bisect a given angle.
¶ Let the given angle which must be divided be angle a.b.c. I will make the lines containing it, a.b. and b.c., equal by the third [proposition], and I will draw line a.c., upon which I will construct an equilateral triangle a.d.c., and I will draw line b.d. I say that this line divides the given angle into equal parts. I understand two triangles, a.b.d. and c.b.d. Since the two sides a.b. and b.d. of triangle a.b.d. are equal to the two sides c.b. and b.d. of triangle c.b.d., and base a.d. is equal to base c.d., therefore by the preceding [proposition], angle a.b.d. is equal to angle c.b.d., which is what was proposed to be done.

To bisect a proposed straight line.
¶ Let the proposed line which must be divided into equal parts be line a.b. Upon it, I will construct an equilateral triangle a.b.c., and I will divide angle c into equal parts according to the teaching of the preceding [proposition] by line c.d. I say that line c.d. divides the given line a.b. into equal parts. For I understand two triangles, a.c.d. and b.c.d., and I argue thus: the two sides a.c. and c.d. of triangle a.c.d. are equal to the two sides b.c. and c.d. of triangle b.c.d., and angle c of one is equal to angle c of the other; therefore, by the fourth [proposition], base a.d. is equal to base d.b., which is what was proposed.