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 CD = CE and\( \angle DBE = \angle BAD\). Show that triangle ACE is isosceles.
+682 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.
