Stuck on this.

In triangle ABC, the angle bisector of $\angle BAC$ meets $\overline{BC}$ at D, such that AD = AB. Line segment $\overline{AD}$ is extended to E, such that CD = CE and $\angle DBE = \angle BAD$. Show that triangle ACE is isosceles. 

What I currently have is:

Since $\overline{AD}$ and $\overline{BD}$ are the same length, then $\bigtriangleup{ABC}$ is isoceles and $\angle{B}$ and $\angle{D}$ are congruent. We let $\angle{ABD}$ equal to $x^\circ$. We let $\angle{B}$ and $\angle{D}$ equal to $y^\circ$. Because $\angle{ADB}$ and $\angle{DCE}$$ are vertical angles, then they are congruent, and therefore $\angle{DCE}$$ is $y^\circ$. $\angle{ADC}$$ and $\angle{BDE}$ are vertical angles as well, and they are both ${180-y}^\circ$.

I don't know what else I can derive, and I'm kind of stuck at this point.