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.
[asy]
pair A = (5, 12);
pair B = (0, 0);
pair C = (5 + 2035/69, 0);
pair D = (10, 0);
pair E = (5/2 + 2035/138, -400/23);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E--C);
draw(D--E);
draw(anglemark(B, A, D, 75));
draw(anglemark(D, A, C, 75));
draw(anglemark(E, B, D, 75));
label("$A$", A, N);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, NE);
label("$E$", E, S);
[/asy]
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.
