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Triangle  $$ABC$$ is drawn so that the angle bisector of $$\angle BAC$$ meets $$\overline{BC}$$ at  $$D$$ and so that triangle $$ABD$$ is an isosceles triangle with $$AB = AD$$. Line segment $$\overline{AD}$$ is extended past $$D$$ to $$E$$ so that triangle $$CDE$$ is isosceles with $$CD = CE,$$ and $$\angle DBE = \angle DAB$$ . Show that triangle $$AEC$$ is isosceles.

Image ------>   https://latex.artofproblemsolving.com/d/3/5/d35b00e81a091116526d86f896d59616cfaa2463.png         (it's super dark)

Jan 8, 2020

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Let  angle  BAD  = y  = angle DAC

And since AB = Ad, then angles ABD  and ADB  are equal

So....let angles  ABD, ADB  =  x

Since angle ADB is vertical to angle CDE, then  angle  CDE  = x

And since CD  = CE,   then  angle CDE  = angle DEC  =  x

So  triangles ABD  and CDE  are similar by AA congruency

So angle DCE  = y

And by the exterior angle theorem, angle ABD  = angle DAC + angle ACD

So

x   = y + angle ACD

x - y   = angle ACD

So  angle ACE  =  angle DCE +  angle ACD

So   angle ACE  = y  + (x - y)    =  x

So   angle  DEC  = x = angle AEC

And angle ACE  = x

So   angle AEC  = angle ACE  means that triangle AEC is isosceles   Jan 8, 2020
edited by CPhill  Jan 8, 2020