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Slovenia — National Math Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Slovenia

A triangle $ABC$ is given with the points $D$ on the side $AB$ , $E$ on the side $BC$ and $F$ on the side $AC$ , such that the line $CD$ is perpendicular to the side $AB$ , the line $DE$ is perpendicular to the side $BC$ and the line $DF$ is perpendicular to the side $AC$ . Prove that the points $A$ , $B$ , $E$ and $F$ lie on the same circle.

Solution

Denote $\angle BAC = \alpha$ . The point $D$ lies on the segment $AB$ , so the angles $\angle BAC$ and $\angle CBA$ cannot be obtuse. If the point $D$ coincides with $A$ or with $B$ , the statement is obviously true. For the remainder of the solution we assume that $D$ lies in the interior of the segment $AB$ .

In this case $\angle FDA = \frac{\pi}{2} - \alpha$ and $\angle FDC = \frac{\pi}{2} - \angle FDA = \alpha$ . In the quadrilateral $DECF$ we have $\angle DEC + \angle CFD = \frac{\pi}{2} + \frac{\pi}{2} = \pi$ , so it is cyclic. This implies that $\angle CEF = \angle CDF = \alpha$ , so $\angle FEB = \pi - \angle FEC = \pi - \alpha$ and $\angle BAF + \angle FEB = \alpha + \pi - \alpha = \pi$ . Hence, $ABEF$ is a cyclic quadrilateral as well.

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