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We have a triangle $\triangle ABC$ such that $AB = BC = 5$ and $AC = 4.$ 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.

Rollingblade  Aug 7, 2018
 #1
avatar+964 
+3

The angle bisector theorem also states that the length of the angle bisector satisfies:

 

 \(\overline{CD}=ab-xy, \frac{a}{x}=\frac{b}{y}.\)

 

Therefore, using the dimensions in the current problem, we reach:

 

\(AC\div CD=AB\div BC\)

\({AD}^2=AC\cdot{A}B-CD\cdot{BD}\)

 

Filling in the numbers, we get:

 

\(\frac4{x}=\frac{5}{5-x}\Rightarrow x=\frac{20}9, 5-x=\frac{25}9\)

\({AD}^2=4\cdot{5}-\frac{20}{9}\cdot\frac{25}9=\frac{1120}{81}\)

 

I hope this helped,

 

Gavin

GYanggg  Aug 7, 2018
 #2
avatar+91160 
+1

Thanks, Gavin !!!!....here's another way using  the Law of  Cosines

 

Here's a pic :

Because AD is an angle bisector, then AC /AB  = CD/BD

So 

4/5 = CD/BD

Then there are 9 parts of BC , then  CD are 4 of them...so CD  = 5 (4/9)  = 20/9

So  BD  = 5 - 20/9 =  45/9 - 20/9  = 25/9

 

 

CD^2  = AD^2  + AC^2  - ( AD * 4) cos DAC

(20/9)^2  =  AD^2  + 4^2  - ( AD * 4) cos DAC

(400/81 - 16 - AD^2) / ( - 4AD) = cosDAC     (1)

 

BD^2 = AD^2 + AB^2  - (AD * 5)cosDAB

(25/9)^2  = AD^2 + 5^2  - (AD * 5)cosDAB

(625/81 - 25 - AD^2) / ( - 5AD) = cos DAB   (2)

 

Since  angle DAC = angle DAB, their cosines are equal...so...we can equate (1)  and (2)

 

(400/81 - 16 - AD^2) / ( - 4AD)  = (625/81 - 25 - AD^2) / ( - 5AD) 

Multiply both sides by  -AD

 

( 400/81 - 16 - AD^2) / 4  = ( 625/81 - 25 - AD^2) / 5

Cross-multiply 

5 ((400/81 - 16 - AD^2) = 4 ( 625/81 - 25 - AD^2)    

Simplify

2000/81 - 80  - 5AD^2  = 2500/81 - 100 - 4AD^2

20 - 500/81 = AD^2

[1620 - 500] / 81  = AD^2

1120/ 81   = AD^2

 

 

cool cool cool

CPhill  Aug 7, 2018
edited by CPhill  Aug 7, 2018

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