Question

Prove that the sides opposite to equal angles of a triangle are equal.

Answer 1

Given: In a $△ABC,\angle B=\angle C$.
To prove: $AB=AC$
Construction: Draw $AD$ perpendicular to $BC$
Proof:
$\angle ADB=\angle ADC={90}^o$ (by construction)
$\angle B=\angle C$ (given)
$AD$ is common side
$\therefore △ADB\equiv △ADC$ (by AAS axiom)
hence, $AB=AC$ (by c.p.c.t.c)
so, the sides opposite to equal angles of a triangle are equal.
This is the converse of Isosceles triangle theorem.