TITLE: Proving XY is perpendicular on :CD
QUESTION [7 upvotes]: In cyclic quadrilateral $ABCD$ the point $E$ is in the middle of $BC$, the perpendicular on $BC$ pass the point $E$ and intersect $AB$ in $X$, and the perpendicular on $AD$ pass the point $E$ and intersect $CD$ in $Y$, what is the proof that $XY$ is perpendicular on $CD$.

I posted this problem  before and i deleted it,because the diagram was not good .

REPLY [2 votes]: $$
\Delta BCX - \text{isosceles}  \Rightarrow \angle BXE =\angle CXE.
$$
$$
\angle ABC + \angle ADC = \pi \Rightarrow \angle ABC =\angle ADY.
$$
$$
\angle DMY=\angle BEX=\frac{\pi}{2} \Rightarrow  \Delta DMY \sim \Delta BEX \Rightarrow 
$$
$$
\Rightarrow \angle CYE = \angle BXE = \angle CXE.
$$
This give us that $CEXY$ inscribed quadrilateral and $\angle XYC + \angle CEX=\pi$.
As $\angle CEX = \frac{\pi}{2} \ \Rightarrow \ \angle XYC = \frac{\pi}{2}$ . 
This is not true if $\angle ABC = \angle DCB $ or $\angle ABC = \frac{\pi}{2}$.

Pictures for cases when $\frac{\pi}{2}> \angle ABC > \angle DCB$;  $\angle ABC > \frac{\pi}{2}$. Just for fun. In this cases $\Delta DMY \sim \Delta BEX$ still true.