# 2014 AMC 8 Problems/Problem 9

## 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)

~Education, the Study of Everything

## Video Solution

https://youtu.be/j5KrHM81HZ8 ~savannahsolver

## 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, $\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}$.

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