Question

In Figure, $AOB$ is a diameter of the circle, $C,D$ and $E$ are any three points on the semi-circle . Find the value of $\angle ACD+\angle BED$.

Answer 1

Let us join $BC$.

Since angle in a semicircle is ${90}^{\circ }$, we have,

$\angle$ $ACB$ = ${90}^{\circ }$

As $BCDE$ is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are supplementary

$\therefore$ $\angle$ $BCD+\angle$ $BED={180}^{\circ }$

Now , adding $\angle$ $ACB$ on both sides , we get

$\Rightarrow \angle BCD+\angle BED+\angle ACB=$ ${180}^{\circ }+\angle ACB$

$\Rightarrow (\angle BCD+\angle ACB)+\angle BED=$ ${180}^{\circ }+{90}^{\circ }$

$\Rightarrow \angle ACD+\angle BED=$ ${270}^{\circ }$

Hence, the required answer is ${270}^{\circ }$