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Help I'm stuck

 

In triangle $ABC$, $D$ lies on segment $\overline{BC}$ such that $\overline{AD}$ is an angle bisector.  If $AB = 2$, $AC = 2$, and $BC = 3$, then find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$.  (Express your answer as a fraction in lowest terms.)

 Oct 12, 2023
edited by wiseowl  Oct 12, 2023
 #1
avatar+128732 
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AD is an angle bisector

 

So

 

AB / BD  = AC / CD         

 

2 / BD  = 2 / CD

 

CD / BD  =   2  / 2

 

CD  / BD  =   1 / 1

 

Therefore  the base of triangle ACD = the base of triangle ABD

And they are both under the same  height

 

So....the ratio of their areas =   1 : 1

 

cool cool cool

 Oct 12, 2023

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