Posted by **Anonymous** on Sunday, June 23, 2013 at 8:18am.

If ABC is a triangle with AB=20,BC=22 and CA=24. Let D lie on BC such that AD is the angle bisector of ∠BAC. What is AD2?

- maths -
**MathMate**, Sunday, June 23, 2013 at 9:43am
I assume the last sentence is

"What is AD?" since AD2 does not mean much.

Hints:

Use cosine rule to find ∠A, in triangle ABC.

A=arccos((20²+24²-22²)/(2*20*24))

=arccos(41/80)

= 59° (approx.)

Then

Area of triangle ABC

=(1/2)AB.AC.sin(A)

=(1/2)*20*24*(11√39)/80

=33√39

Since AD is angle bisector, the altitudes of D to AB equals that to AC.

Therefore AD divides the area of triangle ABC in ratios of the bases AB and AC. Thus

Area of ΔADC=(33√39)*(24/(20+24))

=18√39

Area of ΔADC can also be found by the trigonometric formula

Area of ΔADC

=AD*AC*sin(A/2) ... since AD bisects A

=>

18√39 = AD * 24 *sin(A/2)

=>

AD=3√39/2sin(A/2)=19 (approx.)

- maths -
**Anonymous**, Sunday, June 23, 2013 at 6:20pm
It is asking for the length of AD squared or AD^2.

- maths -
**MathMate**, Sunday, June 23, 2013 at 8:33pm
The solution is given for AD.

Are you able to calculate AD²?

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