If $ X$ is the incentre of $\Delta ABY $ show that $∠CAD=90°$.

Question

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 $∠CAD = 90°.$

Answer 1

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

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

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

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

Answer 2

Look at quadrilateral $XCYD$.

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

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

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

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

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

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

7. 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 $CD$.

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

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

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

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

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

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