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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?2012-08-28
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    As I said before, you have to consider then angle in $B$ a non right angle. Otherwise this is not true.2012-08-28
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    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.2012-08-28
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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".2012-08-28
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    The diagram is much better!2012-08-28
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    Can you prove that $ECYX$ is inscribed in the circle?2012-08-28
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    just in case anybody liked the older picture: http://i.stack.imgur.com/kFBN9.jpg2012-08-28
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    It's not a matter of liking or not: the new diagram is *essentially* different from the original one. The question's been flagged already.2012-08-28
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    What's with all the colons?2012-08-28
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    The only difference between the figures is that the new one is reflected, that is, the line $AB$ is on the right side. Well, line $XC$ is missing also (sorry). But the name of points is the same. Nothing to worry.2012-08-28
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    If E is in the middle of CB, then XE is perpendicular on BC and this contradicts the given that the angle YEB is 90 degrees. Am I missing something?2012-08-28
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    @EmmadKareem YEB is not 90 degrees, XEB is2012-08-28
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    @Sidd, thanks for your reply.2012-08-29
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    I seriously believe that there is a missing constraint or a missing specification in this problem. This because the point Y can be freely positioned and no constraint is given to control where it should lie. as a result, the angle XYC can't be defined using the inputs provided.2012-08-29
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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

1 Answers 1

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enter image description here

$$ \Delta BCX - \text{isosceles} \Rightarrow \angle BXE =\angle CXE. $$

$$ \angle ABC + \angle ADC = \pi \Rightarrow \angle ABC =\angle ADY. $$

$$ \angle DMY=\angle BEX=\frac{\pi}{2} \Rightarrow \Delta DMY \sim \Delta BEX \Rightarrow $$

$$ \Rightarrow \angle CYE = \angle BXE = \angle CXE. $$

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.2012-08-29
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    Ok, I need some time2012-08-29