# 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 $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.$

~Argonauts16 (Diagram by Brendanb4321)

## Solution 2

Alternatively, we could have used similar triangles. We start similarly to 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. 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 know: $$\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.$$

~ alleycat (Diagram by Brendanb4321)

## Solution 3

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

~mn28407

## Solution 4

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

Notice that if <BEC is 90 degrees, then <BCE and <ACE must be 20 degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that <EBD ~ <ECD = 20 degrees. Thus <CBF is 70 - 20 = 50 degrees, and so <BFC is 180 - 20 - 50 = 110 degrees, which is D.

## See Also

 2019 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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