Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$. 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 $\angle BFC$?
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$
Without loss of generality, we assume the length of each side of $ABCD$ is $2$.
Because $E$ is the midpoint of $CD$, $CE = 1$.
Because $ABCD$ is a rhombus, $\angle BCE = 180^\circ - \angle D$.
In $\triangle BCE$, following from the law of sines,
\[ \frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} . \]
We have $\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$.
Hence,
\[ \frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} . \]
By solving this equation, we get $\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}$.
Because $AF \perp BF$,
\begin{align*} BF & = AB \cos \angle ABF \\ & = 2 \cos \left( 46^\circ - \angle FBC \right) . \end{align*}
In $\triangle BFC$, following from the law of sines,
\[ \frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} . \]
Because $\angle BCF = 180^\circ - \angle BFC - \angle FBC$, the equation above can be converted as
\[ \frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} . \]
Therefore,
\begin{align*} \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\ & = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\ & = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\ & = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\ & = - \tan \frac{134^\circ}{2} \\ & = - \tan 67^\circ \\ & = \tan \left( 180^\circ - 67^\circ \right) \\ & = \tan 113^\circ . \end{align*}
Therefore, $\angle BFC = \boxed{\textbf{(D)} \ 113}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)