Let _ABC_ be the triangle, with _AB_ = _AC_. Conceive of this as two triangles; then _AB_ = _AC_, _AC_ = _AB_, and [L]_A_ is common; hence the [triangles]_ABC_ and _ACB_ are congruent, and [L]_B_ of the one equals [L]_C_ of the other. This is a better plan than that followed by some textbook writers of imagining [triangle]_ABC_ taken up and laid down on _itself_. Even to lay it down on its "trace" is more objectionable than the plan of Pappus. 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: [Illustration: =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_.