Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.0 National olympiad, round 2 Prove it

Let $A B C$ be a triangle with $\angle A=90^{\circ}$ and let $D$ be the foot of the altitude from $A$ . The midpoints of $A D$ and $A C$ are called $E$ and $F$ respectively. Let $M$ be the center of the circumcircle of $\triangle B E F$ . Prove that $A C \| B M$ .

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Solution

Due to the right angles at $A$ and $D$ , we have $\triangle A D B \sim \triangle C A B$ (hh). From this, it follows that $\frac{|A D|}{|A B|}=\frac{|C A|}{|C B|}$ . Since $|A E|=\frac{1}{2}|A D|$ and $|C F|=\frac{1}{2}|C A|$ , it also follows that $\frac{|A E|}{|A B|}=\frac{|C F|}{|C B|}$ . From the previous similarity, we also get $\angle B A E=\angle B A D=\angle B C A=\angle B C F$ , so with (zhz) we now have $\triangle A E B \sim \triangle C F B$ . This implies $\angle A B E=\angle C B F$ .
Furthermore, $E F$ is a midline in triangle $A D C$ , so $E F \| B C$ . With Z-angles, this gives $\angle C B F=\angle B F E$ . Due to the inscribed angle theorem and subsequently the angle sum in isosceles triangle $E B M$ , we have $\angle B F E=\frac{1}{2} \angle B M E=90^{\circ}-\angle E B M$ . Therefore, $\angle A B E=\angle C B F=90^{\circ}-\angle E B M$ . We conclude that $\angle A B M=\angle A B E+\angle E B M=90^{\circ}$ . Thus, $A C \| B M$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.