Question

In the figure, ABC is an equilateral triangle. Find $\angle BDCand\angle BEC$.

Answer 1

$\Delta ABC$ is an equilateral triangle.

$\therefore \angle A={60}^o$

$\because \angle BACand\angle BDC$ are in the same segment.

$\therefore \angle BAC=\angle BDC={60}^o$

$\because$ BECD is a cyclic quadrilateral.

$\therefore \angle BDC+\angle BEC={180}^o$

$\Rightarrow {60}^o+\angle BEC={180}^o$

$\Rightarrow \angle BEC={180}^o-{60}^o={120}^o$

Hence, $\angle BDC={60}^{o}and\angle BEC={120}^o$