Question

Angles opposite to equal sides of an isosceles triangle are equal. Prove this result

Answer 1

Consider $\Delta ABC$ an isosceles triangle

$\Rightarrow$ $AB=AC$

Construction: $AD$ is bisector on $BC$

So, $\angle BAD=\angle CAD$

In $△ABD$ and $△ACD$

$\Rightarrow$ $AB=AC$

$\Rightarrow$ $\angle BAD=\angle CAD$

$\Rightarrow$ $AD=AD$

So, $△ABD\cong △ACD$ ..........(SAS)

hence, $\angle ABC=\angle ACB$