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Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

Let the circumcircle of triangle $ABC$ be $\omega$ , and let its excircle tangent to side $BC$ be $\Omega_A$ . Let the intersection points of $\omega$ and $\Omega_A$ be $X$ and $Y$ . Let $P$ be the projection of $A$ onto the tangent line to $\Omega_A$ at point $X$ , and let $Q$ be the projection of $A$ onto the tangent line to $\Omega_A$ at point $Y$ . Let the tangent to the circumcircle of triangle $APX$ at point $P$ , and the tangent to the circumcircle of triangle $AQY$ at point $Q$ , meet at point $R$ .
Prove that line $AR$ and $BC$ are perpendicular to each other.

Solution

Let $D$ be the point of tangency of $BC$ and $\Omega_A$ , and let $D'$ be the antipode of $D$ on $\Omega_A$ . Let $R'$ be the (unique) point satisfying $AR' \perp BC$ and $R'D' \parallel BC$ . We will prove that $R = R'$ .

Let line $PX$ meet $AB$ and $D'R'$ at points $S$ and $T$ , respectively. Let $U$ be the point at infinity where the parallel lines $BC$ and $D'R'$ meet. Since the (degenerate) hexagon $ASXTUC$ is circumscribed about the circle $\Omega_A$ , by Brianchon's theorem, the three lines $AT$ , $SU$ , $XC$ are concurrent at a point, denoted $V$ . Therefore $VS \parallel BC$ . Thus
$\angle(SV, VX) = \angle(BC, CX) = \angle(BA, AX),$
hence $AXSV$ is concyclic. From this we get
$\angle(PX, XA) = \angle(SV, VA) = \angle(R'T, TA).$
Since $\angle APT = \angle AR'T = 90^\circ$ , we know $APR'T$ is concyclic. Therefore
$\begin{aligned} \angle(XA, AP) &= 90^\circ - \angle(PX, XA) = 90^\circ - \angle(R'T, TA) \\ &= \angle(TA, AR') = \angle(TP, PR'). \end{aligned}$
From this we know that $PR'$ is tangent to circle $(APX)$ .

Similarly, $QR'$ is also tangent to circle $(AQY)$ . Therefore $R = R'$ , so $AR \perp BC$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.