In isosceles trapezoid $ABCD$ the median $MN$ cuts diagonals $AC$ and $BD$ at $P$ and $Q$, respectively. If $PQ$ is $4$ find the perimeter of the trapezoid. I found that $AC=BD=8$ and I know that $PQ$ is half of $AB-CD$ but I can't find $CD$. Any help would be appreciated. Angle A is 60 degrees. $MN$ is 8.
Edit: I wasn't given that $MN$ is 8. Now when I know that I can find the perimeter. $P=a+b+2(a-b)=32$