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

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<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>
 
<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>
  
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Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Noting that they subtend to the same arc of <math>180^{\circ}</math>, we can find <math>\angle ECB=20</math> and <math>\angle DBC=50</math> by the triangle angle sum. Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180-50-20=\boxed{(\textbf{\text{D})110}}.</math>
+
Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Noting that they subtend to the same arc of <math>180^{\circ}</math>, we can find <math>\angle ECB=20</math> and <math>\angle DBC=50</math> by the triangle angle sum. Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180-50-20=\boxed{(\textbf{\text{D}})110}.</math>
  
 
~Argonauts16
 
~Argonauts16

Revision as of 18:45, 9 February 2019

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. Noting that they subtend to the same arc of $180^{\circ}$, we can find $\angle ECB=20$ and $\angle DBC=50$ by the triangle angle sum. Then, we take triangle $BFC$, and find $\angle BFC=180-50-20=\boxed{(\textbf{\text{D}})110}.$

~Argonauts16

See Also

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