Question 97
Let $ABCDE$ be a convex pentagon with $BC\parallel AE$ and $\overline{AB}=\overline{AE}$. Let $F$ lie on segment $AE$ such that $\overline{AB}=\overline{BC}+\overline{AF}$ and $\measuredangle CBA=\measuredangle FDC$. Let $M$ be the midpoint of $CF$ and $O$ the circumcenter of triangle $BCD$.
Prove that if $DM\perp MO$, then $\measuredangle FDC=2\cdot\measuredangle ADB$.
Prove that if $DM\perp MO$, then $\measuredangle FDC=2\cdot\measuredangle ADB$.