In figure, AOB is a diameter of the circle

Since, $A, C, D$ and $E$ are four point on a circle, then $A C D E$ is a cyclic quadrilateral.

$\angle A C D+\angle A E D=180^{\circ} \ldots$ (i) $\quad$ [sum of opposite angles in a cyclic quadrilateral is $180^{\circ}$ ]

Now, $\angle A E B=90^{\circ}$ ......(ii)

We know that, diameter subtends a right angle to the circle.

On adding Eqs. (i) and (ii), we get

$(\angle A C D+\angle A E D)+\angle A E B=180^{\circ}+90^{\circ}=270^{\circ}$

$\Rightarrow$ $\angle A C D+\angle B E D=270^{\circ}$