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

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<math>\triangle{ABC}</math> is isosceles so <math>\angle{CAB}=70^{\circ}</math>. Since <math>CB</math> is a diameter, <math>\angle{CDB}=\angle{CEB}=90^{\circ}</math>. Quadrilateral <math>ADFE</math> is cyclic since <math>\angle{ADF}+\angle{AEF}=180^{\circ}</math>. Therefore <math>\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110^{\circ}}</math>
 
<math>\triangle{ABC}</math> is isosceles so <math>\angle{CAB}=70^{\circ}</math>. Since <math>CB</math> is a diameter, <math>\angle{CDB}=\angle{CEB}=90^{\circ}</math>. Quadrilateral <math>ADFE</math> is cyclic since <math>\angle{ADF}+\angle{AEF}=180^{\circ}</math>. Therefore <math>\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110^{\circ}}</math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/O_o_-yjGrOU?t=849
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 +
~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
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Education, the Study of Everything
 
Education, the Study of Everything
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/KXwjFdwrfqk
 +
Includes small notebook concept summary, and where to learn the concepts in longer format.
  
 +
~IceMatrix
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:58, 22 June 2025

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 by OmegaLearn

https://youtu.be/O_o_-yjGrOU?t=849

~ pi_is_3.14

Video Solution

https://youtu.be/GmQIEX4Izt4

Education, the Study of Everything

Video Solution by TheBeautyofMath

https://youtu.be/KXwjFdwrfqk Includes small notebook concept summary, and where to learn the concepts in longer format.

~IceMatrix

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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