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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 $∠CAD = 90^{\circ}$.

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First we show that point $Y$ lies on the edges $CD$. Look at quadrilateral $XCYD$. We will prove that $\angle \, CYD = 180^{\circ}$.

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  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 $CD$.

Now, we are ready to prove that $\angle \, CAD = 90^{\circ}$.

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

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

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

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

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

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

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  • $\begingroup$ Off topic: Nice answer!, but what program or appl. you use for draw this?? Thanks $\endgroup$
    – MathUser
    Nov 6, 2016 at 18:43
  • $\begingroup$ @MathUser I use Inkscape. It is not very fancy, but I am very comfortable with it. A matter of habit I would say. $\endgroup$ Nov 7, 2016 at 2:38

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