Hello there,
The question i have been given is:
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 angle DBE = angle BAD. Show that triangle ACE is isosceles.
Figure image:
I've worked up to this point:
Let's assume x for angles ABD and ADB since they are equal
If they are equal, then angle BAD is equal to 180 - 2x
And since Segment D is bisecting angle A, BAD and CAE are equal, so CAE also equals 180 -2x
angle ADB = CDE = x
I'm not sure what to do up to this point, if someone could push me in the right direction that would be really helpful! Thank you in advance!!
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.
