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Slovenia — National Math Olympiad

Geometry Difficulty 4.5 AIME Prove it Slovenia

Let $ABC$ be an acute triangle. A line parallel to $BC$ intersects the sides $AB$ and $AC$ at $D$ and $E$ . The circumcircle of the triangle $ADE$ intersects the segment $CD$ at $F$ , $F \neq D$ . Prove that the triangles $AFE$ and $CBD$ are similar.

Solution

The lines $DE$ and $BC$ are parallel, so $\angle DCB = \angle CDE$ . The inscribed angles over the chord $EF$ in the cyclic quadrilateral $ADFE$ are equal, $\angle FDE = \angle FAE$ . This implies
$\angle DCB = \angle CDE = \angle FDE = \angle FAE.$
The lines $DE$ and $BC$ are parallel, so $\angle ABC = \angle ADE$ . Since the points $A, D, E$ and $F$ are concyclic, we have $\angle ADE = \angle AFE$ , and so $\angle DBC = \angle EFA$ .
The triangles $AFE$ and $CBD$ have two angles in common, $\angle AFE = \angle DBC$ and $\angle FAE = \angle DCB$ , hence they are similar.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.