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Romanian Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Romania

Consider the isosceles triangle $ABC$ , with $m(\angle BAC) = 100^\circ$ . Let $BD$ be the angle bisector of the angle $\widehat{ABC}$ , with $D \in (AC)$ , the point $E \in BD$ such that $D \in (BE)$ and $BE = BC$ , and the point $F \in (BC)$ such that $AB = BF$ . Prove that the lines $AC$ and $EF$ are orthogonal.

Solution

Triangles $ABD$ and $FBD$ are congruent (S.A.S.), so that $m(\angle FDB) = m(\angle ADB) = 60^\circ$ , and $m(\angle FDC) = 60^\circ$ .

Triangle $EBC$ is isosceles ( $BE = BC$ ), with $m(\angle EBC) = 20^\circ$ , hence $m(\angle BCE) = 80^\circ$ , and from the hypothesis we have $m(\angle ACB) = 40^\circ$ , so $\angle FCD = \angle DCE$ .

Thus, triangles $FCD$ and $ECD$ are congruent (A.S.A.), hence triangle $FCE$ is isosceles. $CD$ is an internal angle bisector, hence also a height. We conclude that $AC \perp FE$ .

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