XX OBM

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