The Teaching of Geometry

by David Eugene Smith

Theorem. If two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isosceles.

The statement is, of course, tautological, the last five words being unnecessary from the mathematical standpoint, but of value at this stage of the student's progress as emphasizing the nature of the triangle. Euclid stated the proposition thus, "If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another." He did not define "subtend," supposing such words to be already understood. This is the first case of a converse proposition in geometry. Heath distinguishes the logical from the geometric converse. The logical converse of Euclid I, 5, would be that "some triangles with two angles equal are isosceles," while the geometric converse is the proposition as stated. Proclus called attention to two forms of converse (and in the course of the work, but not at this time, the teacher may have to do the same): (1) the complete converse, in which that which is given in one becomes that which is to be proved in the other, and vice versa, as in this and the preceding proposition; (2) the partial converse, in which two (or even more) things may be given, and a certain thing is to be proved, the converse being that one (or more) of the preceding things is now given, together with what was to be proved, and the other given thing is now to be proved. Symbolically, if it is given that a = b and c = d, to prove that x = y, the partial converse would have given a = b and x = y, to prove that c = d.

Several proofs for the proposition have been suggested, but a careful examination of all of them shows that the one given below is, all things considered, the best one for pupils beginning geometry and following the sequence laid down in this chapter. It has the sanction of some of the most eminent mathematicians, and while not as satisfactory in some respects as the reductio ad absurdum, mentioned below, it is more satisfactory in most particulars. The proof is as follows:

Given the triangle ABC, with the angle A equal to the angle B.

To prove that AC = BC.

Proof. Suppose the second triangle A'B'C' to be an exact reproduction of the given triangle ABC.

Turn the triangle A'B'C' over and place it upon ABC so that B' shall fall on A and A' shall fall on B.

Then B'A' will coincide with AB.
Since ∠A' = ∠B', Given
and ∠A = ∠A', Hyp.
∴∠A = ∠B'.
∴B'C' will lie along AC.
Similarly, A'C' will lie along BC.