# Prove that CX and CY are perpendicular

There is given convex quadrilateral ABCD. And internal bisectors of angle $\angle A$ and $\angle C$ intersect in point X. And internal bisectors of angle $\angle B$ and $\angle D$ intersect in point Y. And $\angle XAY=90^\circ$. Prove that $\angle XCY = 90^\circ$. Help please!

Credit of the picture goes to user MvG.

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Please respect the attribution requirement of the CC-BY-SA license: when you copy my picture from my answer, you should identify me as its author. –  MvG Dec 4 '12 at 22:02

See the figure below.

Let $b_1$, $b_2$, $b_3$, $b_4$ bisectors of the internal angles of quadrilateral ABCD.

We know that $b_1$ is perpendicular to $v$, then $v$ is the external angle bisector of angle A.

We can deduce the following facts:

1. $Y$ is at the same distance from $s$ and $r$. ($Y \in b_2$).
2. $Y$ is at the same distance from $r$ and $u$. ($Y \in v$).
3. $Y$ is at the same distance from $u$ and $t$. ($Y \in b_4$).

Based on the three facts presented above, we conclude Y is at same distance from $s$ and $t$, and $YC$ bisects the exterior angle $C$. So $b_3$ and $CY$ are perpendicular and therefore $CX$ and $CY$ are perpendicular too.

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@user51650. Have you seen my approach? –  RicardoCruz Jan 4 '13 at 22:17

This comment from Ada gave me the idea for the following solution. Unfortunately, as Ada correctly pointed out later on, the approach is flawed as the angular bisectors of the triangles I mention are not those of the quadrilateral.

The below anser is therefore incorrect and only here for reference.

The three internal angle bisectors of a triangle meet in a point inside the triangle. In a similar fashion, two external and one internal angle bisector will meet in a point outside the triangle.

So consider triangle $ABC$. Take external bisectors for $\angle A$ and $\angle C$, and the internal bisector for $\angle B$. These three lines meet in a point called $Y$. Using $D$ instead of $B$ you can teel that the internal bisector fro $\angle D$ goes through $Y$ as well.

Now you can reverse your argumentation: you know that the point $Y$ is defined by the fact that the two internal bisectors for $\angle B$ and $\angle D$ intersect in this point. So now you can see that the external angle bisectors at $\angle A$ and the line $AY$ have to be one and the same. Likewise for $C$ instead of $A$. So now you know these connecting lines to be external angle bisectors, which of course are orthogonal to the corresponding internal angle bisectors.

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Is this answer correct? Because internal bisector D and B meet in point Y outside of quadrilateral ABCD –  user51650 Dec 5 '12 at 16:03