smc

Since $AD=AE$, we know $EC=AD$ if and only if triangle $ACE$ is isosceles and $\angle ACE=\angle CAE$. Letting $\angle ACE=\theta$, we want to find when $\angle CAE=\theta$. We know $\angle ABC=\angle AEB=2\theta$, so $\angle EAB=180-4\theta$. We also know $\angle CAB=180-3\theta$, and since $\angle CAE=\angle CAB-\angle EAB$, $\angle CAE=\theta$. Since we now know that $\angle CAE=\angle ACE$ regardless of $\theta$, we have $0^{\circ }<\theta <60^{\circ }$, or $\boxed{\text{for all}\theta \text{such that}{0}^{\circ }<\theta <60^{\circ }}$.