Problem

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 $\angle BFC$ ?

Solution (Law of Sines and Law of Cosines)

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 $\angle BCE = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$ .

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)

Solution 2

Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$ .

Because $\overline{AB} \parallel \overline{ED}$ , we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$ , so $\triangle ABG \sim \triangle DEG$ by AA.

Because $ABCD$ is a rhombus, $AB = CD = 2DE$ , so $AG = 2GD$ , meaning that $D$ is a midpoint of segment $\overline{AG}$ .

Now, $\overline{AF} \perp \overline{BE}$ , so $\triangle GFA$ is right and median $FD = AD$ .

So now, because $ABCD$ is a rhombus, $FD = AD = CD$ . This means that there exists a circle from $D$ with radius $AD$ that passes through $F$ , $A$ , and $C$ .

AG is a diameter of this circle because $\angle AFG=90^\circ$ . This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$ , so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$ , which means that $\angle BFC = \boxed{\textbf{(D)} \ 113}$

~popop614

Solution 3

Let $\overline{AC}$ meet $\overline{BD}$ at $O$ , then $AOFB$ is cyclic and $\angle FBO = \angle FAO$ . Also, $AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE$ , so $\frac{AF}{BO} = \frac{AC}{BE}$ , thus $\triangle AFC \sim \triangle BOE$ by SAS, and $\angle OEB = \angle ACF$ , then $\angle CFE = \angle EOC = \angle DAC = 67^\circ$ , and $\angle BFC = \boxed{\textbf{(D)} \ 113}$

~mathfan2020

Solution 4

Observe that all answer choices are close to $112.5 = 90+\frac{45}{2}$ . A quick solve shows that having $\angle D = 90^\circ$ yields $\angle BFC = 135^\circ = 90 + \frac{90}{2}$ , meaning that $\angle BFC$ increases with $\angle D$ . Substituting, $\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}$

~mathfan2020

Solution 5 (Similarity & Circle Geometry)

Let's make a diagram, but extend $AD$ and $BE$ to point $G$ .

$[asy] pair A = (0,0); label("$D$", A, NW); pair B = (2.25,3); label("$A$", B, NW); pair C = (6,3); label("$B$", C, NE); pair D = (3.75,0); label("$C$", D, SE); pair E = (1.875,0); label("$E$", E, S); draw(C--E); draw(A--B--C--D--E--cycle); label("$F$", (3.3,1), S); draw(B--(3.5,1.2)); draw(rightanglemark(B,(3.5,1.2),E)); draw((3.5,1.2)--D); label("$G$", (-2.25, -3), SW); draw(A--(-2.25, -3)); draw(E--(-2.25, -3)); [/asy]$

We know that $AB=2, AD=2, DE=1$ , and $CE=1$ .

By SAS Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$ .

This means that, $2AD=AG$ and $AD \cong DG$ .

$AG=2AD=2(2)=4$ .

This also can prove that $D$ is the midpoint of $AG$ .

Now, let's redraw our previous diagram, but add a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$ , which is on the circle.

$[asy] pair A = (0,0); label("$D$", A, NW); pair B = (2.25,3); label("$A$", B, NW); pair C = (6,3); label("$B$", C, NE); pair D = (3.75,0); label("$C$", D, SE); pair E = (1.875,0); label("$E$", E, S); draw(C--E); draw(A--B--C--D--E--cycle); label("$F$", (3.3,1), S); draw(B--(3.5,1.2)); draw(rightanglemark(B,(3.5,1.2),E)); draw((3.5,1.2)--D); label("$G$", (-2.25, -3), SW); draw(A--(-2.25, -3)); draw(E--(-2.25, -3)); pair O1 = (0,0); draw(circle(O1,3.75)); label("$H$", (-3.75,0), SW); draw(D--(-3.75,0)); [/asy]$

Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with arch $CG$ .

Lets call $\angle CFE = \theta$ .

$\angle FDG$ also intercepts arch $CG$ , but it's vertical angle ( $\angle ADH$ ), also intercepts an arch. So $\angle FDG = 2\angle CFE$ .

$\angle FDG = 2\theta$ .

Notice how $FDG$ and $ADC$ are supplementary to each other.

This concludes that, $2\theta=180-\angle ADC$ .

$2\theta=180-46$

$2\theta=134$

$\theta=67°$ .

Realize how $\angle BFC=180-\theta$

$\angle BFC=180-67$

$\angle BFC= 113.$ Which means the answer is $\boxed{\textbf{(D)} \ 113}$ .

Video Solution

https://youtu.be/Ysb1EK_5B2g

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing

https://youtu.be/lEmCprb20n4

~ pi_is_3.14

2022 AMC 10B (Problems • Answer Key • Resources)
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