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.



Image ------>   https://latex.artofproblemsolving.com/d/3/5/d35b00e81a091116526d86f896d59616cfaa2463.png         (it's super dark)

 Jan 8, 2020

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


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

20 Online Users