Question

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

Answer 1

Since RACP is a cyclic quadrilateral it follows that ∠RPA = ∠RCA =90° - ∠A. Similarly, from cyclic quadrilateral BAQP we get ∠QPA =90° - ∠A. This shows that PH is the angular bisector of ∠RPQ. 

We next show that R, A, Q are collinear. For this, note that ∠BPC = ∠A. Since ∠BHC = 180° - ∠A it follows that BHCP is a cyclic quadrilateral .Therefore ∠RAP + ∠QAP = ∠RCP + ∠QBP = 180°. This proves that R, A, Q are collinear. 

Now ∠QRC = ∠ARC = ∠APC = ∠PAC = ∠PRC. This proves that RC is the angular bisector of ∠PRQ and hence H is the incentre of triangle PQR.