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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 $CD$.

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

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    @EmmadKareem, the most important assumption is that the mediatrix of **$CB$ intersects $AB$** (at $X$), that is, the angle at $B$ is not a right one. Thus, the angle at $D$ is not a right one also. Then, the perpendicular $EM$ for sure will intersect $CD$ at $Y$.2012-08-29

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$ \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 $\angle ABC = \frac{\pi}{2}$.

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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.

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    Ok, I need some time2012-08-29