2007 AMC 12B Problem 7

Because $AB=BC=EA$ and $\angle A=\angle B=90^\circ,$ quadrilateral $ABCE$ is a square, so $\angle AEC=90^\circ$ and $EC$ equals the common side length. Then $CD=DE=EC,$ so $\triangle CDE$ is equilateral and $\angle CED=60^\circ.$ Therefore $$\begin{aligned} \angle E&=\angle AEC+\angle CED \\ &=90^\circ+60^\circ \\ &=150^\circ. \end{aligned}$$ Thus, the correct answer is E.