jmc

Let the measure of $\angle A$ be $x$, so we have $\angle B=x$ and $\angle C=x$, too. Since $\angle A$ is $40^{\circ }$ less than $\angle D$, we have $\angle D=x+40^{\circ }$, so $\angle E=x+40^{\circ }$. The sum of the angle measures in a pentagon is $180(5-2)=540$ degrees, so we have $$x+x+x+(x+40^{\circ })+(x+40^{\circ })=540^{\circ }.$$ Simplifying the left side gives $5x+80^{\circ }=540^{\circ }$, so $5x=460^{\circ }$ and $x=92^{\circ }$. Therefore, $\angle D=\angle A+40^{\circ }=\boxed{132^{\circ }}$.