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.
Diagram is from this question:https://web2.0calc.com/questions/helppp_36
They are not the exact same. The other one says CD and CE are equal. Plz help
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.