2019 AMC 10A Problems/Problem 13

Problem

Let $\Delta ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Contruct 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

$[asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$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 $\triangle DBC$ .

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

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

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

~Argonauts16 (Diagram by Brendanb4321)

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2019 AMC 10A Problems/Problem 13

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