In triangle ABC, the angle bisector of angle BAC meets BC at D, such that AD = AB. Line segment AD is extended to E, such that CD = CE and angle DBE = \angle BAD. Show that triangle ACE is isosceles.
+128475 Since AB = AD, then angle ABD = angle ADB
Let them be = x
So angle BAD = 180 - 2x = angle CAE
And angle CDE is vertical to angle ADB..so it too = x
And since CD = CE, then angle CDE = angle CED = x
So....angle CEA = angle CED = x
So....angle ACE = 180 - angle CAE - angle CEA
So ...angle ACE = 180 - (180 - 2x) - x = 2x - x = x
But this is the same measure as angle CEA
Therefore, in triangle AEC, angle ACE = angle CEA
Therefore...it is isosceles
