2007 AMC 10B Problem 7

Because $AB=BC=EA$ and $\angle A=\angle B=90^\circ,$ quadrilateral $ABCE$ is a square, so $\angle AEC=90^\circ.$ The remaining sides satisfy $CD=DE=EC,$ so $\triangle CDE$ is equilateral and $\angle CED=60^\circ.$ Therefore $\angle E=\angle AEC$ $+\angle CED=90^\circ$ $+60^\circ=150^\circ.$ Thus, the correct answer is E.