Question

In given figure $ABC$ and $DBC$ are two isosceles triangles on the same base $BC$. Show that $\angle ABD=\angle ACD$.

Answer 1

In $△ABC$, we have

$AB=AC$

$\Rightarrow \angle ABC=\angle ACB$... (1) $\mid$ Since angles opposite to equal sides are equal

In $△BCD$, we have

$BD=CD$

$\Rightarrow \angle DBC=\angle DCB$ ... (2) $\mid$ Since angles opposite to equal sides are equal

Adding (1) and (2), we have

$\angle ABC+\angle DBC=\angle ACB+\angle DCB$

$\Rightarrow \angle ABD=\angle ACD$