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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!!

May 12, 2021

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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.

May 12, 2021