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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.

 

 Mar 19, 2020
 #1
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so all we need to prove is that angle aec = angle ace but how?

 Mar 20, 2020
 #2
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someone please help me

 Mar 20, 2020
 #3
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Since  AB  = AD, then angle ABD   = angle ADB   

 

Let them  be  = x

 

So  angle BAD  = 180 - 2x  = angle   CAE

 

And angle CDE  is vertical to angle ADB..so it too   = x

 

And since CD  = CE, then angle  CDE  = angle CED  = x   

 

So....angle CEA   =  angle CED   =   x

 

So....angle  ACE  =  180  - angle  CAE  - angle   CEA   

 

So  ...angle ACE  =  180 - (180 - 2x) - x  =   2x - x  =  x

 

But this is the same measure as angle  CEA

 

Therefore, in triangle AEC, angle ACE = angle CEA

 

Therefore...it is isosceles

 

 

cool cool cool

 Mar 20, 2020
edited by CPhill  Mar 20, 2020

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