# Difference between revisions of "2019 AMC 10A Problems/Problem 13"

## Problem

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Construct the circle with diameter $\overline{BC}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$. What is the degree measure of $\angle BFC ?$

$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

## Solution 1

$[asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("A",(1,0),SE);label("C",(0,2.75),N);label("B",(-1,0),SW);label("E",(0,0),S);label("D",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));[/asy]$

Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$. We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$.

$$\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}$$

$$\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}$$

Then, we take triangle $BFC$, and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}.$

## Solution 2

Alternatively, we could have used similar triangles. We start similarly to Solution 1.

Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Therefore, $$\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.$$

So, $\triangle BEF \sim BDA$ by AA Similarity, since $\angle EBF = \angle DBA$ and $\angle BEC = 90^{\circ} = \angle BDA$. Thus, we know $$\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.$$

Finally, we deduce $$\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.$$

## Solution 3 (outside angles)

Through the property of angles formed by intersecting chords, we find that $$m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}$$

Through the Outside Angles Theorem, we find that $$m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}$$

Adding the two equations gives us $$m\angle BFC + m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB$$

Since $\overarc{BC}$ is the diameter, $m\overarc{BC}=180^{\circ}$, and because $\triangle ABC$ is isosceles and $m\angle ACB=40^{\circ}$, we have $m\angle CAB=70^{\circ}$. Thus $$m\angle BFC=180^{\circ}-70^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}$$

## Solution 4

Notice that if $\angle BEC = 90^{\circ}$, then $\angle BCE$ and $\angle ACE$ must be $20^{\circ}$. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20^{\circ}$. Thus $\angle CBF = 70 - 20 = 50^{\circ}$, and so $\angle BFC = 180 - 20 - 50 = 110^{\circ}$, which is $\boxed{\textbf{(D)}}$.

## Solution 5

$\triangle{ABC}$ is isosceles so $\angle{CAB}=70^{\circ}$. Since $CB$ is a diameter, $\angle{CDB}=\angle{CEB}=90^{\circ}$. Quadrilateral $ADFE$ is cyclic since $\angle{ADF}+\angle{AEF}=180^{\circ}$. Therefore $\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110^{\circ}}$

## Video Solution

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