Question

In the given figure, $ABCDE$ is a pentagon, inscribed in a circle with centre ${}^{\prime }{O}^{\prime }$. $CD=$ $DE$ and $AD=BD.\angle EAD=40^{\circ },\angle BOC=90^{\circ }$. Then, the value of $\angle ADE$ is____

Answer 1

Answer: 45

Since $AD=BD$ and $CD=DE$,
$AE=BC\Rightarrow \angle BOC=\angle AOE=90^{\circ }$
$\Rightarrow \angle ADE=(\frac{1}{2})\angle AOE=(\frac{1}{2})90^{\circ }=45^{\circ }$