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# Help

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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.

Aug 7, 2018

#1
+985
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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

Aug 7, 2018
#2
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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)

(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) / 4  = ( 625/81 - 25 - AD^2) / 5

Cross-multiply

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

Simplify

[1620 - 500] / 81  = AD^2