Triangles $ABC$ and $ADC$ are [isosceles](https://artofproblemsolving.com/wiki/index.php/Isosceles) with $AB=BC$ and $AD=DC$. Point $D$ is inside triangle $ABC$, angle $ABC$ measures 40 degrees, and angle $ADC$ measures 140 degrees. What is the degree measure of angle $BAD$?
$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60$
[2007 AMC12A-6.png](https://artofproblemsolving.com/wiki/index.php/File:2007_AMC12A-6.png)
We angle chase and find out that:

$\angle DAC=\frac{180-140}{2} = 20$
$\angle BAC=\frac{180-40}{2} = 70$
$\angle BAD=\angle BAC- \angle DAC=50\ \mathrm{(D)}$
~minor edits by mobius247