imc

First, we draw an altitude to $BC$ from $E$. Let it intersect at $M$. As $△BEC$ is isosceles, we immediately get $MB=MC=6$, so the altitude is $8$. Now, let $AB=CD=x$. Using the Pythagorean Theorem on $△EMA$, we find $AE=\sqrt{x^2+12x+100}$. From symmetry, $DE=\sqrt{x^2+12x+100}$ as well. Now, we use the fact that the perimeter of $△AED$ is twice the perimeter of $△BEC$. We have $2\sqrt{x^2+12x+100}+2x+12=2(32)$ so $\sqrt{x^2+12x+100}=26-x$. Squaring both sides, we have $x^2+12x+100=676-52x+x^2$ which nicely rearranges into $64x=576\to x=9$. Hence, AB is 9 so our answer is $\boxed{9}$.