In the figure, $\angle A$, $\angle B$, and $\angle C$ are right angles.  If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE$, then $\angle CDE =$

$\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ$
Because $\angle BED=\angle BDE$, $\angle B=90^\circ$, and $\triangle BED$ is a triangle, we get: \[\angle B + \angle BED +\angle BDE=180\] \[90 +\angle BED +\angle BED=180\] \[2\angle BED=90\] \[\angle BED=\angle BDE=45^\circ\]
So $\angle AED=\angle AEB +\angle BED=40 +45=85^\circ$ Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: \[\angle A +\angle C +\angle CDE +\angle AED=360\] \[90 +90 +\angle CDE +85=360\] \[\angle CDE=95^\circ \text{(E)}\]