Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.7 AIME, harder Prove it Brazil

$ABC$ is a triangle. $D$ is the midpoint of $AB$ , $E$ is a point on the side $BC$ such that $BE = 2EC$ and $\angle ADC = \angle BAE$ . Find $\angle BAC$ .

Solution

Let $F$ be the midpoint of $BE$ . Thus $DF$ and $AE$ are parallel. Let $M$ be the intersection point of $AE$ and $CD$ .

$E$ is the midpoint of $CF$ . Since $ME$ is parallel to $DF$ , by Thales theorem $DM = MC$ . But triangle $ADM$ is isosceles, so $AM = DM = MC$ and triangle $AMC$ is isosceles as well. Let $\alpha = \angle ADC = \angle BAE$ . Then $\angle AMC = 2\alpha$ and $\angle MAC = 90^\circ - \alpha$ and, consequently, $\angle BAC = \angle DAM + \angle MAC = \alpha + 90^\circ - \alpha = 90^\circ$ .

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