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Danube Mathematical Competition

Geometry Difficulty 4.5 AIME Prove it Romania

Consider $ABCD$ a rectangle of center $O$ with $AB \ne BC$ . The perpendicular dropped from $O$ to $BD$ intersects lines $AB$ and $BC$ in $E$ and $F$ . Let $M$ and $N$ be the midpoints of segments $[CD]$ and $[AD]$ . Prove that $FM \perp EN$ .

Solution

Let $P$ be the midpoint of $[BC]$ and $Q$ be the intersection point of $EO$ and $CD$ . As $[PM]$ is the midsegment of the triangle $BCD$ , it results that $PM \parallel BD$ and $OQ \perp BD$ , so $QF \perp PM$ . But $PC \perp MQ$ , so $F$ is the orthocenter of the triangle $MPQ$ , which leads to $PQ \perp MF$ . Since the quadrilateral $ENQP$ is a parallelogram, we have $PQ \parallel EN$ . Consequently, $FM \perp EN$ .

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