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China Hong Kong Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Hong Kong

Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\Gamma$ such that $AB = AD$ . Let $E$ be a point on the segment $CD$ such that $BC = DE$ . The line $AE$ intersects $\Gamma$ again at $F$ . The chords $AC$ and $BF$ meet at $M$ . Let $P$ be the symmetric point of $C$ about $M$ . Prove that $PE$ and $BF$ are parallel.

Solution

Since $\angle EFM = \angle AFB = \angle DCA = \angle ECM$ , the points $M$ , $C$ , $F$ , $E$ are concyclic. Using the concyclic points, we obtain
$\angle QME = \angle FME = \angle FCE = \angle FCD = \angle FBD = \angle QBD.$
This implies $EM \parallel DB$ , and hence $\frac{DE}{EQ} = \frac{BM}{MQ}$ . Also, as $\angle MCQ = \angle MCB$ , we have $\frac{BM}{MQ} = \frac{BC}{CQ}$ . Combining these, we obtain $\frac{DE}{EQ} = \frac{BC}{CQ}$ . It follows that $EQ = CQ$ . Now, as $M$ and $Q$ are midpoints of $CP$ and $CE$ , we know that $PE \parallel MQ$ as desired.

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