Japan Mathematical Olympiad

Without loss of generality, we may assume that the points $B$, $C$, $P$ lie on the same straight line in this order. Let us denote by $Q^{\prime }$ the point of intersection of the lines $AB$ and $PQ$ and by $R^{\prime }$ the point of intersection of the lines $AC$ and $PR$. If $\angle ACB\ne 90^{\circ }$, we see from $P{Q}^{\prime }\perp A{Q}^{\prime }$ and $P{R}^{\prime }\perp A{R}^{\prime }$ that the 4 points $A$, $Q^{\prime }$, $P$, $R^{\prime }$ lie on the circumference of a circle. Therefore, if we let $S$ be the point of intersection of the lines $BC$ and $Q^{\prime }{R}^{\prime }$, then we get $$\begin{array}{rl}\angle BS{Q}^{\prime }& =\angle BP{Q}^{\prime }+\angle P{Q}^{\prime }S\\ & =\angle BP{Q}^{\prime }+\angle PA{R}^{\prime }\\ & =\angle BP{Q}^{\prime }+\angle PB{Q}^{\prime }\\ & =\angle B{Q}^{\prime }Q=90^{\circ },\end{array}$$ which means that $BC\perp Q^{\prime }{R}^{\prime }$. If $\angle ACB=90^{\circ }$, we see that the points $Q^{\prime }$ and $R^{\prime }$ coincide with the points $A$ and $C$, respectively, and therefore, $BC\perp Q^{\prime }{R}^{\prime }$ holds in this case as well. $Q^{\prime }$, $R^{\prime }$ are the mid-points of the line segments $PQ$ and $PR$, respectively, and therefore, we must have $QR\parallel Q^{\prime }{R}^{\prime }$, and thus we have shown that $BC\perp QR$ must hold.