Question 132
Consider a convex quadrilateral $ABCD$ whose diagonals are perpendicular and meet at $E$. Let $P \in [AD]$ be a point such that $P \neq A$ and $PE = EC$. Suppose that the circumcircle of triangle $BCD$ meets $[AD]$ again at a point $Q$ different from $A$. Let $R$ be the second intersection point of $[AC]$ with the circle passing through $A$ and tangent to $(EP)$ at $P$. Suppose that the points $B, R, Q$ are collinear. Show that $\widehat{BCD} = 90^{\circ}$.