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# geometry problem

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We have a triangle ABC such that AB=6, BC=8,  and CA=$$12$$.  If AD is an angle bisector such that D is on BC,  then find the value of AD^2.

Feb 9, 2022

#1
+689
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ABC has lengths 6, 8, 12. To find angle B, we can use law of cosines.

12^2 = 8^2 + 6^2 - 2*8*6*cos(angB)

144 = 64 + 36 - 96cosB.

-11/24 = cosB.

Now that we have cosB, we can use the Angle Bisector theorem to get length of BD. 8/12 = BD/DC, so BD is 12/5.

Then we can apply the law of cosines again.

AD^2 = 8^2 + (12/5)^2 - 2*12/5*8*cos(angB)

AD^2 = 64 + 144/25 - (-88/5)

AD^2 = 1600/25 + 144/25 + 440/25

$${AD}^2 = {2184\over25}$$

$$AD^2 = 87.36$$

Feb 9, 2022
#2
+1696
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You got this wrong:   8/12 = BD/DC

Should be:  6/12 = BD/DC        (But there's nothing you can do with this.)

The correct way is          6/12 = BD / (8 - BD)        correct!!!

civonamzuk  Feb 10, 2022
#3
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## BD = 2 2/3

Who did give you all those ++++?????

Guest Feb 16, 2022
#4
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I guess... all 4 moderators gave him/her ++++

Guest Feb 17, 2022