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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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Is this homework? –  VF1 Aug 28 '12 at 0:58
    
As I said before, you have to consider then angle in $B$ a non right angle. Otherwise this is not true. –  Sigur Aug 28 '12 at 1:01
    
When you deleted the question, that also had the effect of deleting the comments people had taken the trouble to make. Better to have just edited the new diagram into the original question. –  Gerry Myerson Aug 28 '12 at 1:06
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@sigur There is nothing wrong with what he said: if one of the hypotheses is that XE and BA meet at X, then angle B simply isn't a right angle. You don't have to "exclude that case". –  rschwieb Aug 28 '12 at 1:22
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Can you prove that $ECYX$ is inscribed in the circle? –  Sigur Aug 28 '12 at 1:31

1 Answer 1

up vote 5 down vote accepted

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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.

enter image description here

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Please, could you change the colors of the lines? It is too light. –  Sigur Aug 29 '12 at 12:59
    
Ok, I need some time –  Mike Aug 29 '12 at 13:00

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