+239 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.
+983 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
+129852 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
