Converse Of Isosceles Triangle Theorem

Theorem: Sides opposite to the equal angles in a triangle are equal.

Prerequisites:
AAS congruency (proof)

Proof:

Let ABC be a triangle having $\angle B = \angle C$. Let us draw AD which bisects the $\angle A$ and meets BC at D.

In $\triangle ABD$ and $\triangle ACD$,

$\qquad\quad\angle B = \angle C\qquad\qquad\qquad\;\:\!\:\!\qquad\text{(given)}\\ \qquad\quad\angle BAD = \angle CAD\qquad\qquad\quad\:\text{(by construction)}$
Also, $\quad AD = AD\qquad\qquad\qquad\qquad\text{(common)}$

Hence, $\triangle ABD\cong\triangle ACD$ by AAS rule.

Thus, by CPCTC,  AB = AC,  i.e,  $\triangle ABC$ is an isosceles triangle.

Q.E.D.


Recommended:
Isosceles Triangle Theorem
Angle Bisector Theorem
Location of centroid