Question

In a triangle $ABC,$ let $H$ denote its orthocentre. Let $P$ be the reflection of $A$ with respect to $BC.$ The circumcircle of $△ABP$ intersects the line $BH$ again at $Q,$ and the circumcircle of $△ACP$ intersects the line $CH$ again at $R.$ Prove that $H$ is the incentre of $△PQR.$

Answer 1

$RACP$ is a cyclic quadrilateral
$\therefore \angle RPA=\angle RCA={90}^o-\angle A$
Similarly, $BAQP$ is a cyclic quadrilateral, $\therefore \angle QPA={90}^o-\angle A$
$\therefore PH$ is the angular bisector of $\angle RPQ$.
$\angle BPC=\angle A$................Since $\angle BHC={180}^o-\angle A$ it follows that BHCP is a cyclic quadrilateral
$\therefore \angle RAP+\angle QAP=\angle RCP+\angle QBP={180}^o$
Therefore R, A, Q are collinear.
Now $\angle QRC=\angle ARC=\angle APC=\angle PAC=\angle PRC$
$\therefore RC$ is the angular bisector of $\angle PRQ$
$\therefore H$ is the incenter of $△PQR$