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Slovenia — National Math Olympiad, 2012

Geometry Difficulty 4.5 AIME Prove it Slovenia

Let a point $E$ lie on the side $CD$ of a square $ABCD$ . Let a point $F$ lie on the line $AB$ , but not on the line segment $AB$ , and let $|BF| = |DE|$ . Prove that the lines $AC$ and $EF$ are perpendicular.

Solution

Let $T$ be the intersection point of the lines $AC$ and $EF$ . Because $|BF| = |DE|$ , $|BC| = |DA|$ and $\angle CBF = \angle ADE = \frac{\pi}{2}$ , the triangles $ADE$ and $CBF$ are congruent. We thus have $|AE| = |CF|$ , and the quadrilateral $AFCE$ is an isosceles trapezoid. Hence $\angle EFA = \angle CAF$ , which is also equal to $\frac{\pi}{4}$ because $AC$ is a diagonal of the square $ABCD$ . We get $\angle ATF = \pi - \angle EFA - \angle CAF = \frac{\pi}{2}$ , from which we conclude that the lines $AC$ and $EF$ are perpendicular.

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