Brazilian Mathematical Olympiad

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, $\angle BAC = \angle DAM + \angle MAC = \alpha + 90^\circ - \alpha = 90^\circ$.