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BxMO Team Selection Test

Geometry Difficulty 7.2 National Olympiad, round 2 Prove it Netherlands

Let $ABCD$ be a cyclic quadrilateral with $|AB| = |BC|$ . Point $E$ lies on the arc $CD$ which does not contain $A$ and $B$ . The intersection of $BE$ and $CD$ is denoted by $P$ , the intersection of $AE$ and $BD$ is denoted by $Q$ . Prove that $PQ \parallel AC$ .

Solution

Because $|AB| = |BC|$ , we have $\angle AEB = \angle BDC$ , hence $\angle QEP = \angle AEB = \angle BDC = \angle QDP$ , which yields that $QPED$ is a cyclic quadrilateral. Therefore, $\angle QPD = \angle QED = \angle AED = \angle ACD$ . From this, we get that $QP$ and $AC$ are parallel. $\square$

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