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$
Solution
Take $G$ as the midpoint of $AB$ and connect $DG$, meeting $AF$ at $H$.
$DE\parallel GB$ and $DE=GB$, so $DGBE$ is a parallelogram and $DG\parallel BE$; hence $DG\bot AF$.
In triangle $AFB$, $HG\parallel BF$ and $G$ is the midpoint of $AB$, so $H$ is the midpoint of $AF$.
So $DH$ is the perpendicular bisector of $AF$, and $DF=DA=DC$.
Triangles $DFC$ and $DFA$ are isosceles, so $\angle DFC=\angle DCF$ and $\angle DFA=\angle DAF$. Their angle sums give $\angle DCF+\angle DFC+\angle DFA+\angle DAF=360^\circ-\angle ADC=314^\circ$. Since $\angle AFB=90^\circ$, $\angle BFC=360^\circ-90^\circ -\angle CFD-\angle DFA=270^\circ - \frac12\cdot314^\circ=113^\circ$.
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 $\angle{BFC}$?
