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Team Selection Test for JBMO

Geometry Difficulty 4.8 AIME Prove it Turkey

In a convex quadrilateral $ABCD$ , the diagonals intersect at $E$ , $BE = \sqrt{2} \cdot ED$ and $\angle BEC = 45^\circ$ . Let $F$ be the foot of the perpendicular from $A$ to $BC$ and $P$ be the second intersection point of the circumcircle of triangle $BFD$ and the line segment $[DC]$ . Find $\angle APD$ .

Solution

Let $Q$ be the foot of the perpendicular from $B$ to the line $AC$ . Then we obtain $BQ = EQ = ED$ which means that $\angle AQD = 22.5^\circ$ . Since the points $B, F, Q, A$ are concyclic, we have $CF \cdot CB = CQ \cdot CA$ . Since the points $B, F, P, D$ are also concyclic we have $CF \cdot CB = CP \cdot CD$ and hence $CQ \cdot CA = CP \cdot CD$ and therefore the points $A, Q, P, D$ are concyclic. So, $\angle APD = \angle AQD = 22.5^\circ$ .

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