Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Romania

In a circle, consider two chords $[AB]$ , $[CD]$ that intersect at $E$ . The lines $AC$ and $BD$ meet at $F$ . Let $G$ be the projection of $E$ onto $AC$ . We denote by $M, N, K$ the midpoints of the segment lines $[EF]$ , $[EA]$ , and $[AD]$ , respectively. Prove that the points $M, N, K, G$ are concyclic.

Marius Bocanu

Solution

Let $P$ be the midpoint of $[AF]$ ; points $N, G, P, M$ are on the Euler circle of triangle $AEF$ , which means they are concyclic $(*)$ .

As $NK \parallel DE$ and $NM \parallel AF$ , we have:
$\begin{align*} \angle KNM &= \angle KNE + \angle ENM = \angle DEB + \angle BAC \\ &= \angle DEB + \angle BDE = \angle ABF, \end{align*}$
and, as $KP \parallel DF$ and $MP \parallel AE$ , we have
$\angle KPM = 180^\circ - \angle KPA - \angle MPF = 180^\circ - \angle DFA - \angle EAF = \angle ABF.$
It follows that the quadrilateral $KNPM$ is cyclic, hence $K, N, P, M$ are concyclic. Combining this with $(*)$ gives the conclusion.

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