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.

Guest Feb 9, 2022

#1**0 **

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

Thus, AD^2 is:

\({AD}^2 = {2184\over25} \)

\(AD^2 = 87.36\)

proyaop Feb 9, 2022

#2**0 **

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