Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Romania

Let $AA'$ , $BB'$ , $CC'$ be the altitudes from the vertices of an acute-angled triangle $ABC$ . Points $E$ and $F$ lie on the segments $CB'$ and $BC'$ respectively, such that
$B'E \cdot C'F = BF \cdot CE.$
Prove that the quadrilateral $AEA'F$ is cyclic.

Solution

The given relation writes as $\frac{BF}{FC'} = \frac{B'E}{EC}$ . Let $M$ be a point on the side $BC$ such that $FM \parallel CC'$ . Then $\frac{BM}{MC} = \frac{BF}{FC'} = \frac{B'E}{EC}$ , implying $ME \parallel BB'$ . Hence the angles $\angle AEM$ , $\angle AFM$ and $\angle AA'M$ are right angles, therefore $E$ , $F$ , $A'$ all lie on the circle of diameter $AM$ . The conclusion follows.

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