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)
+128475 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
