1975 AMC 12 Problem 24

Because $AB=AE=AD,$ first study the isosceles triangle $ABE$

Whether $E$ lies on $BC$ or its extension, angle chasing gives $\angle EAC=\angle ECA=\theta$

Since $AB=AE,$ triangle $ABE$ is isosceles. For $0^\circ\lt\theta\lt45^\circ,$ $E$ lies between $B,C$; the exterior-angle theorem in triangle $AEC$ gives $2\theta=\angle EAC+\theta,$ so $\angle EAC=\theta.$ For $45^\circ\lt\theta\lt60^\circ,$ $E$ lies beyond $B$; then $\angle AEC=180^\circ-2\theta,$ hence $$\begin{aligned} \angle EAC &=180^\circ\\ &\quad-(180^\circ-2\theta)-\theta\\ &=\theta. \end{aligned}$$ At $\theta=45^\circ,\ E=B$ and the conclusion is immediate. Thus triangle $AEC$ is always isosceles, so $EC=EA=AD.$

Therefore, the correct answer is E.