jmc

We first draw a diagram: Since $△ABC$ is isosceles, $AD$ is a median as well as an altitude. We then see that $△DEC\sim △BAC$ and since $D$ is the midpoint of $BC,$ the area of $△DEC$ is $\frac{1}{4}$ of the area of $△ABC,$ or 45. The area of $ABDE$ is, therefore, $180-45=\boxed{135}.$