Here is the direct theorem: proof of isosceles triangle?
Is posible to prove the reciprocal of that theorem that it's:
In one triangle with two congruent angles it oppossed two congruent sides.
I know a proof but it uses the fact that the sum of the measure of the angles of any triangle is equal to 180° and using the criteria Angle-Side-Angle: just bisect the remaining angle, and apply that result in both triangle.
But I'm not very happy with that, there exist another proof "more elementary"?
Best Answer
You have a triangle $\triangle(AA'C)$ with equal angles at $A$ and $A'$. Then $\triangle(AA'C)$ is congruent to $\triangle(A'AC)$ because the angles at $A$ and $A'$ are equal, and $|AA'|=|A'A|$. It follows that $|AC|=|A'C|$.