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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
avatar+370 
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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°    indecision

 

  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.      wink      

 

 

 

      

 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
 #3
avatar+107524 
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Why doesn't it?

Melody  Feb 5, 2020

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