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# Stuck on this.

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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.

Jun 17, 2020
edited by gwenspooner85  Jun 17, 2020

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This is AoPS homework.  Stop trying to cheat on your homework.

Jun 17, 2020
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@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.

gwenspooner85  Jun 17, 2020