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.
Use cosine rule to find ∠A, in triangle ABC.
= 59° (approx.)
Area of triangle ABC
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))
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)
It is asking for the length of AD squared or AD^2.
The solution is given for AD.
Are you able to calculate AD²?