2022 AMC 10B Problem 20

Extend $BE$ beyond $E$ to meet line $AD$ at $G$. Because $BC \parallel AD$ and $E$ is the midpoint of $CD$, triangles $BCE$ and $GDE$ are congruent (two equal alternate angles and $CE = DE$). Hence $DG = BC = AD$, so $D$ is the midpoint of $AG$.

Triangle $AFG$ has a right angle at $F$, and $FD$ is the median to its hypotenuse, so $DF = DA = DG$. The rhombus also gives $DC = DA$. Therefore $A$, $F$, $G$ and $C$ all lie on the circle with center $D$ and radius $DA$.

$F$ lies on segment $BE$, which is on the same side of line $CG$ as $A$ (both $B$ and $E$ are), so $F$ and $A$ are on the same arc $CG$ and the inscribed angles agree:
$$\angle CFG = \angle CAG = \angle CAD.$$
The diagonal $AC$ bisects the rhombus angle at $A$, and $\angle BAD = 180^\circ - 46^\circ = 134^\circ$, so $\angle CAD = 67^\circ$.

Finally $B$, $F$, $G$ are collinear with $F$ between $B$ and $G$, so
$$\angle BFC = 180^\circ - \angle CFG = 180^\circ - 67^\circ = 113^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 113}$.

A midpoint on one side of a parallelogram plus a line through the opposite vertex almost always begs to be extended: the congruent triangles it creates double a segment and produce a new midpoint. The right angle at $F$ then converts that midpoint into a circle center (median to the hypotenuse), and the equal sides of the rhombus place one more point on the same circle. Angles that looked inaccessible become inscribed angles.

• Trying to compute the angle by brute-force trigonometry or coordinates, which is slow and error-prone without a calculator.
• Stopping after $DF = DA = DG$ and not noticing that $DC$ is also a radius; without $C$ on the circle the angle at $F$ cannot be reached.