Answer: E

Small Hint:

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ$

Big Hint:

$\angle AED = \angle AEB + \angle BED;$ then use quadrilateral $AEDC,$ whose angles sum to $360^\circ$

Solution:

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ,$ so $\angle BED = \angle BDE = 45^\circ.$

Then $\angle AED = \angle AEB + \angle BED$ $= 40^\circ + 45^\circ$ $= 85^\circ.$ In quadrilateral $AEDC,$ the angles at $A$ and $C$ are $90^\circ,$ so

$$\begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ. \end{aligned}$$

Thus, the correct answer is E .