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

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==Problem==
 
==Problem==
  
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Let <math>\Delta ABC</math> be an isosceles triangle with <math>BC = AC</math> and <math>\angle ACB = 40^{\circ}</math>. Contruct the circle with diameter <math>\overline{BC}</math>, and let <math>D</math> and <math>E</math> be the other intersection points of the circle with the sides <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively. Let <math>F</math> be the intersection of the diagonals of the quadrilateral <math>BCDE</math>. What is the degree measure of <math>\angle BFC ?</math>
+
Let <math>\triangle ABC</math> be an isosceles triangle with <math>BC = AC</math> and <math>\angle ACB = 40^{\circ}</math>. Construct the circle with diameter <math>\overline{BC}</math>, and let <math>D</math> and <math>E</math> be the other intersection points of the circle with the sides <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively. Let <math>F</math> be the intersection of the diagonals of the quadrilateral <math>BCDE</math>. What is the degree measure of <math>\angle BFC ?</math>
  
 
<math>\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</math>
 
<math>\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</math>
  
−
==Solution==
+
==Solution 1==
−
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{(\text{D})110}.</math>
+
 
 +
<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 <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find <math>\angle ABC=70^{\circ}</math>. We can find <math>\angle ECB=20^{\circ}</math> and <math>\angle DBC=50^{\circ}</math> by the triangle angle sum on <math>\triangle ECB</math> and <math>\triangle DBC</math>.  
 +
 
 +
<cmath>\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}</cmath>
 +
 
 +
<cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}</cmath>
 +
 
 +
Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}.</math>
 +
 
 +
==Solution 2==
 +
 
 +
Alternatively, we could have used similar triangles. We start similarly to Solution 1.
 +
 
 +
Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Therefore, <cmath>\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.</cmath>
 +
 
 +
So, <math>\triangle BEF \sim BDA</math> by AA Similarity, since <math>\angle EBF = \angle DBA</math> and <math>\angle BEC = 90^{\circ} = \angle BDA</math>.
 +
Thus, we know <cmath>\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.</cmath>
 +
 
 +
Finally, we deduce <cmath>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.</cmath>
 +
 
 +
==Solution 3 (outside angles)==
 +
 
 +
Through the property of angles formed by intersecting chords, we find that
 +
<cmath>m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}</cmath>
 +
 
 +
Through the Outside Angles Theorem, we find that
 +
<cmath>m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}</cmath>
 +
 
 +
Adding the two equations gives us
 +
<cmath>m\angle BFC + m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB</cmath>
 +
 
 +
Since <math>\overarc{BC}</math> is the diameter, <math>m\overarc{BC}=180^{\circ}</math>, and because <math>\triangle ABC</math> is isosceles and <math>m\angle ACB=40^{\circ}</math>, we have <math>m\angle CAB=70^{\circ}</math>. Thus
 +
<cmath>m\angle BFC=180^{\circ}-70^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}</cmath>
 +
 
 +
==Solution 4==
 +
 
 +
Notice that if <math>\angle BEC = 90^{\circ}</math>, then <math>\angle BCE</math> and <math>\angle ACE</math> must be <math>20^{\circ}</math>. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that <math>\angle EBD \cong \angle ECD = 20^{\circ}</math>. Thus <math>\angle CBF = 70 - 20 = 50^{\circ}</math>, and so <math>\angle BFC = 180 - 20 - 50 = 110^{\circ}</math>, which is <math>\boxed{\textbf{(D)}}</math>.
 +
 
 +
==Solution 5==
 +
 
 +
<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>
 +
 
 +
== 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
  
 
==See Also==
 
==See Also==

Latest revision as of 10:15, 4 November 2022

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

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
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Problem 12 		Followed by
Problem 14
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All AMC 10 Problems and Solutions

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