Let $ABCD$ be a rhombus with $\angle{ADC} = 52^{\circ}$. Let $E$ be the midpoint of $\overline{CD}$, and let $F$ be the point on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$. What is the degree measure of $\angle{BFC}$?

1. The argument of the problem works for any angle: $DF=DA=DC$, and $\angle BFC=270^\circ-\frac12(360^\circ-\angle ADC)=90^\circ+\frac12\angle ADC$.
2. With $\angle ADC=52^\circ$: $\angle BFC=90^\circ+26^\circ=116^\circ$.