maths
posted by Anonymous .
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?

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+24²22²)/(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.) 
It is asking for the length of AD squared or AD^2.

The solution is given for AD.
Are you able to calculate AD²?
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