+0  
 
0
46
1
avatar

In triangle $ABC$, the side lengths are $AB = BC = 5$ and $AC = 6.$ If $AD$ is an angle bisector such that $D$ is on $BC,$ then find the value of $AD^2.$ Express your answer as a common fraction.

 Aug 10, 2023
 #1
avatar+129270 
+1

 

 

 

Angle Bisector Theorem

 

BD /AB = DC/ AC            let BD = x ......so DC = BC - x =  5 - x

 

x/ 5 = (5-x) / 6

 

6x = 5(5-x)

 

6x = 25 - 5x

 

11x = 25

 

x = 25/11  = BD

 

Law of Cosines

 

AC^2  = AB^2 + BC^2 - 2(AB * BC) cos ABC

 

6^2 = 5^2 + 5^2 - 2(5 * 5)cos ABC

 

[36 - 50 ] / [ -50]  = cos ABC

 

cosABC =  -14/-50  = 7/25

 

Law of Cosines again

 

AD^2 =  AB^2 + BD^2 - 2(AB * BC) cos ABC

 

AD^2 = 5^2 + (5/11)^2 - 2 ( 5 * 5/11) (7/25)

 

AD^2 = 25 + 25/121 - (50/11)(7/25)

 

AD^2 = 2896 / 121

 

cool cool cool

 Aug 10, 2023

3 Online Users

avatar
avatar