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avatar+279 

We have a triangle ABC such that AB = 6, BC = 8, and CA = 10. If AD is an angle bisector such that D is on BC, then find the value of AD^2.

 

I already tried inputting 52 beacuse AD is the hypotenuse of right triangle ABD and I used the pythagorean theorem (6^2 + 4^2 = AD^2) and found the that AD^2 is 52, but I input this answer and it was incorrect :(

 Jul 3, 2019
edited by ANotSmartPerson  Jul 3, 2019
 #1
avatar+105195 
+3

See the following  image : 

 

 

Let  A  = (0,6)     B  = (0,0)     C = (8,0)

 

If AD is an angle bisector, we have by the angle bisector theorem that

 

AB / AC  =  BD / DC

 

6 / 10  =  BD /DC

 

3 / 5  = BD/DC

 

Then  BC  is  divided in 8 equal parts  and BD is 3 of them...so

 

(3/8) BC  =  (3/8)*8  =  3

 

So....BD  =  3     which implies that   D  = (3,0)

 

So  ...by the distance formula

 

AD^2  =  (3 - 0)^2  +  ( 6 - 0)^2

 

AD^2  =  3^2 + 6^2

 

AD^2   =  9 + 36

 

AD^2  =  45

 

 

cool cool cool

 Jul 3, 2019
edited by CPhill  Jul 3, 2019
 #2
avatar+279 
+1

Thanks Phil!

 Jul 4, 2019

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