Question

In figure $\Delta ABC$ is an isosceles triangle with $AB=AC$and $\angle ABC={50}^{\circ }$ . Find $\angle BDC$ and $\angle BEC$.

Answer 1

In $\Delta ABC$, we have
AB = AC
$\Rightarrow \angle ACB=\angle ABC$
$\Rightarrow \angle ACB={50}^{\circ }[\angle ABC={50}^{\circ }]$
$\therefore \angle BAC={180}^{\circ }–(\angle ABC+\angle ACB)$
$\Rightarrow \angle BAC={180}^{\circ }–({50}^{\circ }+{50}^{\circ })={80}^{\circ }$
Since$\angle BAC$ and $\angle BDC$ are angles in the same segment.
$\therefore \angle BDC=\angle BAC\Rightarrow \angle BDC={80}^{\circ }$
Now, BDCE is a cyclic quadrilateral.
$\therefore \angle BDC+\angle BEC={180}^{\circ }$
$\Rightarrow {80}^{\circ }+\angle BEC={180}^{\circ }\Rightarrow \angle BEC={100}^{\circ }$
Hence, $\angle BDC={80}^{\circ }and\angle BEC={100}^{\circ }$