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Difference between revisions of "2014 AMC 8 Problems/Problem 9"
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<math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math>
 
<math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math>
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/jLnqUOe0HPE
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~Education, the Study of Everything
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==Video Solution==
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https://www.youtube.com/watch?v=HP-lBKohxhE  ~David
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https://youtu.be/j5KrHM81HZ8 ~savannahsolver
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==Video Solution ==
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https://youtu.be/abSgjn4Qs34?t=3140
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==Solution==
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Using angle chasing is a good way to solve this problem. <math>BD = DC</math>, so <math>\angle DBC = \angle DCB = 70</math>, because it is an isosceles triangle. Then <math>\angle CDB = 180-(70+70) = 40</math>. Since <math>\angle ADB</math> and <math>\angle BDC</math> are supplementary, <math>\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=8|num-a=10}}
 
{{AMC8 box|year=2014|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:32, 16 June 2024

Problem

In $\bigtriangleup ABC$, $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$. What is the degree measure of $\angle ADB$?

[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35));[/asy]

$\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$

Video Solution (CREATIVE THINKING)

https://youtu.be/jLnqUOe0HPE

~Education, the Study of Everything


Video Solution

https://www.youtube.com/watch?v=HP-lBKohxhE ~David

https://youtu.be/j5KrHM81HZ8 ~savannahsolver

Video Solution

https://youtu.be/abSgjn4Qs34?t=3140

Solution

Using angle chasing is a good way to solve this problem. $BD = DC$, so $\angle DBC = \angle DCB = 70$, because it is an isosceles triangle. Then $\angle CDB = 180-(70+70) = 40$. Since $\angle ADB$ and $\angle BDC$ are supplementary, $\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}$.

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 AJHSME/AMC 8 Problems and Solutions

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