2022 AMC 10B Problem 20

The midpoint $E$ is the clue: extending a line through the midpoint of a side of a rhombus usually creates a pair of equal triangles.

1. Extend $BE$. Continue $BE$ past $E$ until it meets line $AD$ (extended past $D$) at a point $G$.
2. Two equal triangles. Triangles $BCE$ and $GDE$ have $CE = DE$ (since $E$ is the midpoint), equal angles at $E$ (vertical angles), and equal angles at $C$ and $D$ (because $BC$ is parallel to $AD$). So they are congruent, and $DG = CB$.
3. $D$ is a midpoint. All sides of the rhombus are equal, so $DG = CB = AD$. So $D$ is the midpoint of $AG$.
4. $F$ is on a circle around $D$. Triangle $AFG$ has a right angle at $F$. In a right triangle, the midpoint of the longest side is the same distance from all three corners. So $DF = DA = DG$.
5. So is $C$. $DC = DA$ because they are sides of the rhombus. So $A$, $F$, $G$ and $C$ all lie on the circle with center $D$ and radius $DA$.
6. Angle at the center. $G$ lies on line $AD$ past $D$, so $\angle CDG = 180^\circ - \angle ADC = 180^\circ - 46^\circ = 134^\circ$.
7. Angle at $F$. An angle whose corner is on the circle is half the angle at the center that looks at the same arc. The points $C$ and $G$ are seen from $F$, on the far side of the circle, so
$$\angle CFG = \tfrac12 \times 134^\circ = 67^\circ.$$
8. Finish. $B$, $F$, $G$ lie on one straight line with $F$ between $B$ and $G$, so
$$\angle BFC = 180^\circ - \angle CFG = 180^\circ - 67^\circ = 113^\circ.$$

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

Two classic moves work together here. Extending a line through the midpoint of a side makes congruent triangles and a new midpoint. Then a right angle at $F$ turns that midpoint into the center of a circle, because the midpoint of a right triangle's longest side is equally far from all three corners. Once four points sit on a circle, an awkward angle becomes half of an easy one.

Step 7 can also be done by comparing with the corner $A$. The points $A$ and $F$ are both on the circle and on the same side of the chord $CG$, so they see $C$ and $G$ at the same angle: $\angle CFG = \angle CAG$. The diagonal $AC$ cuts the rhombus's angle at $A$ in half, and that angle is $180^\circ - 46^\circ = 134^\circ$. So $\angle CAG = 67^\circ$, and again $\angle BFC = 180^\circ - 67^\circ = 113^\circ$.

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 $\boxed{\textbf{(D)}\ 113}$.