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+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.)
It is asking for the length of AD squared or AD^2.
The solution is given for AD.
Are you able to calculate AD²?
To find AD², we need to apply the angle bisector theorem. According to the angle bisector theorem, the ratio of the lengths of the sides of a triangle is equal to the ratio of the lengths of the corresponding angle bisectors.
In this case, let's assume BD = x and DC = 22 - x (since BD + DC = BC). We can use the angle bisector theorem to find the ratio of AD to AC and then use that ratio to find AD².
The angle bisector theorem states:
AD / AC = BD / BC
Now, substitute the given values:
AD / 24 = x / 22
To solve for x, cross-multiply:
AD = (24 * x) / 22
Now, we can use the fact that AB + BD = AD and BC - DC = BD to find x:
AB + BD = AD
20 + x = (24 * x) / 22
Cross-multiply:
22 * (20 + x) = 24 * x
Simplify:
440 + 22x = 24x
Rearrange to find x:
2x = 440
x = 220
Now that we have x, we can substitute it back into the equation AD = (24 * x) / 22 to find AD:
AD = (24 * 220) / 22
AD = 240
Finally, to find AD²:
AD² = 240²
AD² = 57600
Therefore, AD² is equal to 57600.