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In triangle \(ABC\) ,  \(AB=AC\) and \(D\) is a point on \(\overline{AC}\) so that \(\overline{BD}\) bisects angle \(ABC\). If \(BD=BC\), what is the measure, in degrees, of angle \(A\)?

tertre  Mar 14, 2018
 #1
avatar+91297 
+2

Since AB  = AC, then

Angle ABC  = angle ACB

 

And since BD = BC

Then Angle BDC  = Angle DCB

But DCB  = ACB

 

So....triangles   ABC  and  BDC  are similar

 

Call angle ABC  = 2theta

But since ABC is bisected, then  DBC  is also  = theta

And angle DBC  = angle BAC  = theta

 

Then...in triangle ABC

Angle ABC + Angle ACB  + Angle BAC  = 180

2theta + 2theta + theta  = 180

5theta  = 180

theta  = 36°=  BAC   = "Angle A "

 

 

cool cool cool

CPhill  Mar 14, 2018
 #2
avatar+3282 
+2

I understood it much better now! Thanks so much, CPhill! 

tertre  Mar 14, 2018

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