+4 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 $\angle DBE = \angle BAD$.
Show that triangle $ACE$ is isosceles.
+142 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
