Question

In the figure AB = AC and DB = DC. Prove that $\angle ABD=\angle ACD$.

Answer 1

In $△ABC$, we have

$AB=AC\Rightarrow \angle ABC=\angle ACB$

(angles opposite to equal sides.)

Again in $△DBC$, we have

$DB=DC$ ......... (given)

$\Rightarrow \angle DBC=\angle DCB$ (angles opposite to equal sides).

Hence we obtain

$\angle ABC-\angle DBC=\angle ACB-\angle DCB$.

This gives

$\angle ABD=\angle ACD$. $\left[henceproved\right]$