Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Brazil

Let $ABCD$ be a parallelogram and $\Gamma$ the circumcircle of the triangle $ABD$ . Lines $BC$ and $CD$ meet $\Gamma$ at $E \ne B$ and $F \ne D$ respectively. Prove the circumcenter of the triangle $CEF$ lies on $\Gamma$ .

Solution

Consider all angles oriented and modulo $180^\circ$ . Let $O$ be the center of the circle. In the cyclic pentagon $ABEDF$ , $\angle EBF = \angle EDF = \angle EDC = \angle CED + \angle DCE = \angle BED + \angle DCE = \angle BAD + \angle DCE = 2\angle DCE$ . This means that if $M$ is the midpoint of the arc $FE$ that does not contain $A$ , $\angle FME = \angle EBF = 2\angle DCE = 2\angle FCE$ . Since $ME = MF$ , $M$ is the circumcenter of triangle $CEF$ .

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