Triangle ABC is drawn so that the angle bisector of angle BAC meets BC at D and so that triangle ABD is an isosceles triangle with AB = AD. Line segment 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.
Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.
Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.
Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.