Question

Prove that angles opposite to equal sides of an isosceles triangle are equal.

Answer 1

Here, In $∆ABC$, $AC=BC$.

Let us draw $CD$ which is the bisector of $\angle ACB$.

Therefore, $CD$ divides $∆ABC$ into two parts, $∆ACD$ and $∆BCD$.

In $∆ACD$ and $∆BCD$ we have,

$AC=BC$……(Given)

$\angle ACD=\angle BCD$……(By construction)

$CD=CD$……….(common side)

Thus, $∆ACD\cong ∆BCD$ by S.A.S congruence criterion.

So, $\angle CAD=\angle CBD$……(By C.P.C.T)

Hence, it is proved that angles opposite to equal sides of an isosceles triangle are equal.