Difference between revisions of "2014 AMC 8 Problems/Problem 9"

Revision as of 23:37, 31 December 2022

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

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

https://youtu.be/j5KrHM81HZ8 ~savannahsolver

Video Solution by OmegaLearn

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

~ pi_is_3.14

Solution

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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