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In triangle $ABC$, $\angle ABC = 90^\circ$, and point $D$ lies on segment $BC$ such that $AD$ is an angle bisector.  If $AB = 60$ and $BD = 40$, then find $AC$.

 Oct 12, 2023
 #1
avatar+128794 
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Since AD is  an angle bisector

 

BD / AB  = DC / AC

40 / 60  =  DC / AC

2 / 3  =  DC / AC

(2/3)AC = DC

 

So  using the Pythagorean Theorem

 

AB^2  + BC^2  =  AC^2

 

60^2  +  (BD + DC)^2 =  AC^2

 

60^2  +  ( 40 +  (2/3 * AC) )^2  = AC^2

 

3600 +  1600 + (160/3)AC + (4/9)AC^2  = AC^2

 

(5/9)AC^2   - (160/3)AC  - 5200  =  0

 

Multiply through by  9

 

5AC^2  -  480AC  - 46800  = 0

 

AC  =    [ 480  + sqrt [ 480^2  - 4 * 5 * -46800   ]  /  10    =  

 

[ 480  +  sqrt [1166400]  ]   / 10  = 

 

[ 480 + 1080 ] / 10   =

 

1560  / 10   =

 

156

 

 

cool cool cool

 Oct 12, 2023

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