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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.

 

 

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 Jan 8, 2020
 #1
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Let  angle  BAD  = y  = angle DAC

And since AB = Ad, then angles ABD  and ADB  are equal

So....let angles  ABD, ADB  =  x

 

Since angle ADB is vertical to angle CDE, then  angle  CDE  = x

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

So  triangles ABD  and CDE  are similar by AA congruency

So angle DCE  = y

 

And by the exterior angle theorem, angle ABD  = angle DAC + angle ACD

So

x   = y + angle ACD

x - y   = angle ACD

 

So  angle ACE  =  angle DCE +  angle ACD

So   angle ACE  = y  + (x - y)    =  x

 

So   angle  DEC  = x = angle AEC

And angle ACE  = x

 

So   angle AEC  = angle ACE  means that triangle AEC is isosceles

 

cool cool cool

 Jan 8, 2020
edited by CPhill  Jan 8, 2020

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