Quadrilateral $ABCD$ has $AB = BC = CD$, $\angle ABC = 80^\circ$ and $\angle BCD = 160^\circ$. What is the measure of $\angle BAD$?

1. Put $B=(0,0)$, $C=(1,0)$ and $AB=BC=CD=1$. Then $A=(\cos80^\circ,\sin80^\circ)$ and, since $\angle BCD=160^\circ$, $D=(1+\cos20^\circ,\sin20^\circ)$.
2. $\overrightarrow{AD}=(1+\cos20^\circ-\cos80^\circ,\ \sin20^\circ-\sin80^\circ)=(1+\sin50^\circ,\ -\cos50^\circ)=(1+\cos40^\circ,\ -\sin40^\circ)=2\cos20^\circ\,(\cos20^\circ,\ -\sin20^\circ)$, which points in the direction $-20^\circ$.
3. $\overrightarrow{AB}=(-\cos80^\circ,-\sin80^\circ)$ points in the direction $260^\circ=-100^\circ$.
4. So $\angle BAD=-20^\circ-(-100^\circ)=\boxed{80^\circ}$.