# Cyclic quadrilateral with angle bisectors

Let $ABCD$ be a cyclic quadrilateral. Lines $AB$ and $CD$ meet at point $X$, and lines $AD, BC$ meet at point $Y$. Furthermore let $e$ be the angle bisector of $\angle AYB$, and $AB\cap e = Z$. Let $f$ be the angle bisector of $\angle AID$, where $I$ is the incenter of $\triangle XAD$. Prove that the angle bisector of $\angle AYZ$ is perpendicular to $f$. I tried angle hunting, but it didn't help. Maybe with polar or power of a point?

It is not difficult to show by angle chasing that the angle bisectors of $\widehat{DYC}$ and $\widehat{AXD}$ are orthogonal. The angle bisector of $\widehat{AXD}$ goes through $I$, hence $f$ is given by a rotation of the previous angle bisector with respect to $I$. The angle bisector of $\widehat{AYZ}$ is given by a rotation of the angle bisector of $\widehat{DYC}$ with respect to $Y$. These rotations have the same amplitude (always by angle chasing) hence $f$ is orthogonal to the angle bisector of $\widehat{AYZ}$ as wanted.

I think angle chasing may still be possible:

Let the two lines that we want to show orthogonal meet at $H$. Let the angle bisector of $\angle XDA$ meet $YH$ at $J$.

We know $\angle DYJ=\frac{\angle DYZ}{2}=\frac{\angle DYC}{4}$. Therefore $$\angle DJH=\frac{\angle DYC}{4}+\frac{\angle YDC}{2}$$

On the other hand, $$\angle DIH=\frac{\angle DIA}{2}=\frac{90^{\circ}+\frac{\angle AXD}{2}}{2}=45^{\circ}+\frac{\angle AXD}{4}$$

Therefore, $$\angle DJH+\angle DIH=\frac{\angle DYC}{4}+\frac{\angle YDC}{2}+45^{\circ}+\frac{\angle AXD}{4}$$

Now, we know that $$\angle AXD+\angle DYC+2\angle YDC=\angle AXD+\angle XDA+\angle DYC+\angle YDC=(180^{\circ}-\angle XAD)+(180^{\circ}-\angle YCD)=360^{\circ}-180^{\circ}=180^{\circ}$$

$$\angle DJH+\angle DIH=\frac{180^{\circ}}{4}+45^{\circ}=90^{\circ}$$ so $\angle IHJ=90^{\circ}$ as desired.