+283 We have a triangle ABC such that AB = 6, BC = 8, and CA = 10. If AD is an angle bisector such that D is on BC, then find the value of AD^2.
I already tried inputting 52 beacuse AD is the hypotenuse of right triangle ABD and I used the pythagorean theorem (6^2 + 4^2 = AD^2) and found the that AD^2 is 52, but I input this answer and it was incorrect :(
+128474 See the following image :
Let A = (0,6) B = (0,0) C = (8,0)
If AD is an angle bisector, we have by the angle bisector theorem that
AB / AC = BD / DC
6 / 10 = BD /DC
3 / 5 = BD/DC
Then BC is divided in 8 equal parts and BD is 3 of them...so
(3/8) BC = (3/8)*8 = 3
So....BD = 3 which implies that D = (3,0)
So ...by the distance formula
AD^2 = (3 - 0)^2 + ( 6 - 0)^2
AD^2 = 3^2 + 6^2
AD^2 = 9 + 36
AD^2 = 45
