Question

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 $\angle BFC?$

$\ 110$

$\ 111$

$\ 112$

$\ 113$

$\ 114$

Solution:

First, we extend $BE$ and $AD$ such that they meet at $G.$ Since $\angle GDE = \angle ECB,$ $\angle GED = \angle BEC \text{ and } DE = EC,$ we know $GDE \cong BCE.$ Therefore, $DG =BC = AD.$ This means that if we construct a circle with center $D$ that includes $A,$ $C,G$ are also on it.

Also, since $AFG$ is a right triangle, the drawn circle would be its circumcircle, placing $F$ on the circle.

Since $\angle GDC = 134^\circ,$ we can get $\angle CFE = \angle CFG = \dfrac {\overset{\Large\frown}{CG}} 2$ $= \dfrac{134^\circ}{2} = 67^\circ.$ Therefore, $\angle BFC = 180^\circ - 67^\circ = 113^\circ.$

Thus, the answer is D .

2022 AMC 10B Problem 20

Solution:

Problem 20 in Other Years