Find the sum of angle

$\begin{aligned} \angle AED + \angle DCB & = \blue{\angle AED + \angle DCA} + \red{\angle ACB} & \small \blue{\text{Opposites angles of cyclic quad ACDE}} \\ & = \blue{180^\circ} + \red{90^\circ} & \small \red{\text{Angle extended by diameter.}} \\ & = \boxed{270}^\circ \end{aligned}$

Let $\angle {BmC}=α$ .

Then $\angle {mAE}=\angle {AEm}=\dfrac{3α}{2}, \angle {mED}=\angle {mDE}=\angle {mDC}=\angle {mCD}=\angle {mCB}=\angle {mBC}=\dfrac{π-α}{2}\implies \angle {AED}+\angle {BCD}=\dfrac{π-α}{2}+\dfrac{3α}{2}+π-α=\dfrac{3π}{2}=\boxed {270\degree}$