Slovenija 2008

The midpoint of the hypotenuse is also the circumcentre, so $P$ is the circumcentre of the triangle $AH{B}^{\prime }$ and $|AP|=|PH|=|P{B}^{\prime }|$. The triangle $HP{B}^{\prime }$ is isosceles with the apex at $P$. Let $Q{B}^{\prime }B=\alpha$. Then $\alpha =\angle P{B}^{\prime }H=\angle B^{\prime }HP=\angle BH{A}^{\prime }$. Since $\angle H{A}^{\prime }B+\angle B{C}^{\prime }H=\frac{\pi }{2}+\frac{\pi }{2}=\pi$, points $A^{\prime }$, $B$, $C^{\prime }$ and $H$ are concyclic and $\angle B{C}^{\prime }{A}^{\prime }=\angle BH{A}^{\prime }=\alpha$. In the quadrilateral $B^{\prime }Q{C}^{\prime }R$ we have



$$\angle R{C}^{\prime }Q+\angle Q{B}^{\prime }R=\pi -\angle B{C}^{\prime }A+\angle Q{B}^{\prime }B=\pi -\alpha +\alpha =\pi ,$$ so $B^{\prime }$, $Q$, $C^{\prime }$ and $R$ are concyclic.

Since $A{C}^{\prime }{B}^{\prime }H$ is a cyclic quadrilateral we have $\alpha =\angle B^{\prime }HA=\angle B^{\prime }{C}^{\prime }A$. The connection between the inscribed angles of a cyclic quadrilateral $B^{\prime }Q{C}^{\prime }R$ gives us the equalities $\angle B^{\prime }RQ=\angle B^{\prime }{C}^{\prime }Q=\angle B^{\prime }{C}^{\prime }A=\alpha$. So, $\angle B^{\prime }RQ=\alpha =\angle B^{\prime }HA$ and $AH$ and $QR$ are parallel. Since $AH$ is perpendicular to $BC$ it is also perpendicular to $QR$.