2022 AMC 10B Problem 20

Extend $\overline{BE}$ to meet line $AD$ at $G.$ Because $AD\parallel BC,$ we have $\angle GDE=\angle ECB,$ and $\angle GED=\angle BEC$ are vertical angles. Also $DE=EC,$ so $\triangle GDE\cong\triangle BCE.$ Hence $DG=BC=AD.$ Thus $D$ is the midpoint of $\overline{AG}.$ The circle centered at $D$ through $A$ also passes through $C$ and $G.$ Since $AF\perp FG,$ Thales’ theorem places $F$ on this circle as well. Because $DG$ is opposite $DA,$ $$\begin{aligned}\angle GDC&=180^\circ-\angle ADC\\&=134^\circ.\end{aligned}$$ The inscribed angle $\angle GFC$ subtending arc $GC$ is therefore $67^\circ.$ Finally, $B,F,G$ are collinear, so $$\angle BFC=180^\circ-67^\circ=113^\circ.$$ Thus, the answer is D .