If $ X$ is the incentre of $\Delta ABY $ show that $∠CAD=90°$. In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$. If $X$ is the incentre of triangle $ABY$ , show that $∠CAD = 90°.$
 A: *

*Since the center of an incircle is the intersection of the triangle's bisectors, and since the angles of a triangle sum to $180^\circ$,
$$
 \angle XYB + \angle YBX + \angle XAY = 90^\circ.
 $$

*Since inscribed angles subtended on the same arc are congruent,
$$
 \angle XAY = \angle XDY.
 $$

*Since the angles of a triangle sum to $180^\circ$,
$$
 \angle DYX = 180^\circ - \angle XYB - \underset{=\angle YBX}{\underbrace{\angle YBD}} - \underset{=\angle XDY}{\underbrace{\angle BDY}} = 90^\circ
 $$

*Since $\angle DYX$ and $\angle DAX$ are opposite angles of the cyclic quadrilateral $ADYX$, they add up to $180^\circ$, so, by step 3,
$$
 \angle CAD = \angle DAX = 90^\circ.
 $$

A: Look at quadrilateral $XCYD$.  


*

*$\angle\, XDY = \angle \, XAY = \alpha$ as inscribed in a circle.

*$\angle \, BAC = \angle \, BAX = \angle \, XAY = \alpha$ since $AC$ passes through the incenter $X$ of triangle $ABY$ and therefore $AC$ is the interior angle bisector of angle $\angle \, BAY$.

*$\angle \, XDC = \angle \, BDC = \angle \, BAC = \alpha$ as inscribed in a circle.

*Hence $\angle \, XDC = \angle \, XDY = \alpha$. 

*Analogously, $\angle \, XCD = \angle \, XCY = \beta$.

*In triangle $CDX$ angles sum up to $180^{\circ}$ so  $$180^{\circ} = \angle \, CXD + \angle \, XCD + \angle \, XDC = \angle \, CXD + \alpha + \beta$$ and thus $\angle \, CXD = 180^{\circ} - \alpha - \beta$.

*In quadrilateral $XCDY$ angles sum up to $360^{\circ}$ so $$360^{\circ} = \angle \, CXD + \angle \, XCY + \angle \, XDY + \angle \, CYD  = \angle \, CXD + \alpha + \beta + \angle \, CYD = $$ $$= 180^{\circ} +  \angle \, CYD $$ therefore $\angle \, CYD = 180^{\circ}$ and thus $Y$ lies on $CD$.

*$\angle \, AXD = \angle \, BXC$ since $X$ is the intersection point of $AC$ and $BD$.

*$\angle \, AYD = \angle \, AXD$ and $\angle \, BYC = \angle \, BXC$ as inscribed in corresponding circles. Hence $\angle \, AYD = \angle \, BYC$

*$\angle \, XYA = \angle \, XYB$ since $YX$ is angle bisector of $\angle \, AYC$. Hence
$$\angle \, XYD = \angle \, XYA + \angle \, AYD = \angle \, XYB + \angle \, BYC = \angle \, XYC$$

*But $180^{\circ} = \angle \, XYD + \angle \, XYC = 2 \, \angle \, XYD $ so $\angle \, XYD = 90^{\circ}$.

*As quadrilateral $AXYD$ is inscribed in a circle, $180^{\circ} = \angle \, XAD + \angle \, XYD = \angle \, XAD + 90^{\circ}$ so $\angle \, XAD = 90^{\circ}$.

*$X \in AC$ so $\angle \, CAD = \angle \, XAD = 90^{\circ}$. 
