The First Six Books of the Elements of Euclid

by Euclid; John Casey

Demonstrate this Proposition directly by cutting off from BC a part equal to EF.

PROP. XXVI.—Theorem.

If two triangles (ABC, DEF) have two angles (B, C) of one equal respectively to two angles (E, F) of the other, and a side of one equal to a side similarly placed with respect to the equal angles of the other, the triangles are equal in every respect.

Dem.—This Proposition breaks up into two according as the sides given to be equal are the sides adjacent to the equal angles, namely BC and EF, or those opposite equal angles.

1. Let the equal sides be BC and EF; then if DE be not equal to AB, suppose GE to be equal to it. Join GF; then the triangles ABC, GEF have the sides AB, BC of one respectively equal to the sides GE, EF of the other, and the angle ABC equal to the angle GEF (hyp.); therefore [iv.] the angle ACB is equal to the angle GFE; but the angle ACB is (hyp.) equal to DFE; hence GFE is equal to DFE—a part equal to the whole, which is absurd; therefore AB and DE are not unequal, that is, they are equal. Consequently the triangles ABC, DEF have the sides AB, BC of one respectively equal to the sides DE, EF of the other; and the contained angles ABC and DEF equal; therefore [iv.] AC is equal to DF, and the angle BAC is equal to the angle EDF.

2. Let the sides given to be equal be AB and DE; it is required to prove that BC is equal to EF, and AC to DF. If BC be not equal to EF, suppose BG to be equal to it. Join AG. Then the triangles ABG, DEF have the two sides AB, BG of one respectively equal to the two sides DE, EF of the other, and the angle ABG equal to the angle DEF; therefore [iv.] the angle AGB is equal to DFE; but the angle ACB is equal to DFE (hyp.). Hence (Axiom i.) the angle AGB is equal to ACB, that is, the exterior angle of the triangle ACG is equal to the interior and non-adjacent angle, which [xvi.] is impossible. Hence BC must be equal to EF, and the same as in 1, AC is equal to DF, and the angle BAC is equal to the angle EDF.

This Proposition, together with iv. and viii., includes all the cases of the congruence of two triangles. Part I. may be proved immediately by superposition. For it is evident if ABC be applied to DEF, so that the point B shall coincide with E, and the line BC with EF, since BC is equal to EF, the point C shall coincide with F; and since the angles B, C are respectively equal to the angles E, F, the lines BA, CA shall coincide with ED and FD. Hence the triangles are congruent.

Def.—If every point on a geometrical figure satisfies an assigned condition, that figure is called the locus of the point satisfying the condition. Thus, for example, a circle is the locus of a point whose distance from the centre is equal to its radius.