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!

a busy cat

Credit of the picture goes to user MvG.

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    $\begingroup$ 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. $\endgroup$ – MvG Dec 4 '12 at 22:02

Consider the line XY and try to show that AXCY are concyclic.

Yet more elegantly, observe that AY is the external bisector of A. This should be very helpful to solve the question.

  • $\begingroup$ yea i thought about that. i also tried to prove that (N is middle point of YD) NX = NC which is equual to that AXCY are concyclic but i failed :/ $\endgroup$ – user51650 Dec 5 '12 at 14:15
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    $\begingroup$ I've also realized that today AY is the external bisector of A, so kinda remember a theorem which states internal bisectors of opposite sides and external bisectors of the other sides are concurrent. Yet could not find it :( Still I'm looking for $\endgroup$ – Ada Dec 5 '12 at 14:20
  • $\begingroup$ Great comment there, @Ada. I just managed to phrase an answer based on this. I hope you don't mind me using this crucial idea in this fashion. As far as I am concerned, you are welcome to write your own formulation of this approach, so that you can get your deserved share of the reputation. $\endgroup$ – MvG Dec 5 '12 at 15:37
  • $\begingroup$ I've also considered this as a solution yet when we consider the triangle ABC, the angles A and C change and AY is not external bisector of A anymore, I've stucked at this point of the proof. What do you think about this? $\endgroup$ – Ada Dec 5 '12 at 16:16

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.

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

See the figure below.

Bisectors of a quadrilateral

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.

  • $\begingroup$ @user51650. Have you seen my approach? $\endgroup$ – RicardoCruz Jan 4 '13 at 22:17

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