# 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 $\triangle DBC$.

$\angle{BDC}+\angle{DCB}+\angle{DBC}=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle{DBC}\implies\angleDBC=50^{\circ}$ (Error compiling LaTeX. ! Undefined control sequence.)

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

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

~Argonauts16 (Diagram by Brendanb4321)