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Junior Turkish Mathematical Olympiad

Geometry Difficulty 4.1 AIME Prove it Turkey

In a convex quadrilateral $ABCD$ , the diagonals intersect at the point $E$ and $\angle AEB = 90^\circ$ . A point $P$ is chosen on the side $[AD]$ other than $A$ so that $PE = EC$ . The circumcircle of the triangle $BCD$ intersects the side $[AD]$ at the point $Q$ other than $A$ . The circle passing through $A$ and tangent to the line $EP$ at $P$ intersects the line segment $[AC]$ at the point $R$ . If $B, Q, R$ are collinear, then show that $\angle BCD = 90^\circ$ .

Solution

Let $\angle EAP = \alpha$ and $\angle ECD = \beta$ . Note that $\angle RPE = \alpha$ . Choose a point $X$ on $[ED]$ such that $\angle RPX = 90^\circ$ . Observe that $R, P, X, E$ are cyclic and hence $\angle RXE = \alpha$ which implies that $A, R, X, D$ are concyclic. Therefore $ER \cdot EA = EX \cdot ED$ . On the other hand we have $ER \cdot EA = EP^2 = EC^2$ . Thus we get $EX \cdot ED = EC^2$ which implies that $\angle EXC = \beta$ . Clearly $\angle QDC = 180^\circ - \alpha - \beta$ and hence $\angle QBC = \angle RBC = \alpha + \beta$ since $Q, B, C, D$ are cyclic and $B, Q, R$ are collinear. On the other hand $\angle RXC = \alpha + \beta$ as well and therefore $X$ is the reflection of $B$ with respect to $E$ . Thus we get $\angle EBC = \angle EXC = \beta$ and the result follows.

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