+781 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.
+781 @Guest,
I'm not trying to cheat, I'm simply asking for some help. I even wrote what I had so far. Cheating would just be copy and pasting the question here with no further explanation.
