Let $ABCD$ be a rhombus with $\angle{ADC}=46^{\circ}$, and 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 $\angle{BFC}$?
Draw the diagonals of rhombus $ABCD$, $\overline{AC}$ and $\overline{BD}$, with $\overline{AC}$ intersecting $\overline{BD}$ at $G$, $\overline{BD}$ intersecting $\overline{AF}$ at $H$, and $\overline{AC}$ intersecting $\overline{BE}$ at $I$. Then draw line $\overline{FG}$, and $\overline{EG}$. We have the following diagram:
Proposition 1: $ABFG$ is cyclic quadrilateral, because both $\triangle{ABG}$ and $\triangle{ABF}$ are right triangles, as $\overline{AG}\perp\overline{BG}$ and $\overline{AF}\perp\overline{BF}$. Therefore $$\angle{AFG}=\angle{ABG}=\angle{ADG}=\dfrac{1}{2}\angle{ADC}=23^{\circ},\ \angle{FAG}=\angle{FBG}$$ $$\angle{GFI}=\angle{HFI}-\angle{HFG}=90^{\circ}-23^{\circ}=67^{\circ}$$
Proposition 2: $CEGF$ is cyclic quadrilateral, because $$\angle{GFE}=\angle{GFI}=67^{\circ}=90^{\circ}-23^{\circ}=\angle{CGD}-\angle{CDG}=\angle{GCD}=\angle{GCE}$$ and $$\angle{FGC}=\angle{AFG}+\angle{FAG}=23^{\circ}+\angle{FBG}=\angle{BDE}+\angle{EBD}=\angle{BEC}=\angle{FEC}$$
Because $E$ is midpoint of $CD$ and $\triangle{CDG}$ is a right triangle, we have $$\overline{CE}=\overline{GE}=\overline{ED}$$ Therefore $\triangle{CEG}$ is an isosceles triangle, with $\angle{CGE}=\angle{GCE}=67^{\circ}$.
Because $CEGF$ is cyclic quadrilateral, we have $\angle{CFE}=\angle{CGE}=67^{\circ}$. Therefore $\angle{BFC}=180^{\circ}-\angle{CFE}=180^{\circ}-67^{\circ}=113^{\circ}$. So the answer is $\boxed{(D)}$.
Note: We can also approve $CEGF$ is cyclic as $\triangle{FIG}\sim\triangle{CIE}$, and $\dfrac{\overline{FI}}{\overline{GI}}=\dfrac{\overline{CI}}{\overline{EI}}$, i.e. $\overline{CI}\cdot\overline{GI}=\overline{EI}\cdot\overline{EF}$.
