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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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    Please respect the attribution requirement of the [CC-BY-SA license](http://creativecommons.org/licenses/by-sa/3.0/): when you copy my picture from [my answer](http://math.stackexchange.com/a/250727/35416), you should identify me as its author.2012-12-04

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