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# Help ASAP

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Triangle $$ABC$$ is drawn so that the angle bisector of $$\angle{BAC}$$ meets $$\overline{BC}$$ at $$D$$ and so that triangle $$ABD$$ is an isosceles triangle with $$AB=AD$$. Line segment $$\overline{AD}$$ is extended past $$D$$ to $$E$$ so that triangle $$CDE$$ is isosceles with $$CD = CE$$ and $$\angle DBE = \angle DAB$$. Show that triangle $$AEC$$ is isosceles.

Jan 29, 2020

#1
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There are 4 isosceles triangles:  ABD,  CDE,   BCE,  and   AEC.

Δ ABD  =>   45° + 67.5° + 67.5° = 180°

Δ CDE  =>   45° + 67.5° + 67.5° = 180°

Δ BCE  =>    45° + 90° + 45° = 180°

Δ AEC  =>    45° + 67.5° + 67.5° = 180°

Triangle   AEC  is an isosceles triangle, because 2 of its 3 angles are equal, and as a result of that, 2 of its sides are equal as well.

Jan 29, 2020
edited by Dragan  Jan 30, 2020
edited by Dragan  Jan 30, 2020
#2
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Sorry, but this doesn't explain why

Guest Jan 29, 2020
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Why doesn't it?

Melody  Feb 5, 2020