2
$\begingroup$

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 $f$. ![enter image description here I tried angle hunting, but it didn't help. Maybe with polar or power of a point?

$\endgroup$
0
$\begingroup$

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.

enter image description here

$\endgroup$
0
$\begingroup$

Diagram

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.