+1679 In triangle $ABC$, $\angle ABC = 90^\circ$, and point $D$ lies on segment $BC$ such that $AD$ is an angle bisector. If $AB = 60$ and $BD = 40$, then find $AC$.
+128656 Since AD is an angle bisector
BD / AB = DC / AC
40 / 60 = DC / AC
2 / 3 = DC / AC
(2/3)AC = DC
So using the Pythagorean Theorem
AB^2 + BC^2 = AC^2
60^2 + (BD + DC)^2 = AC^2
60^2 + ( 40 + (2/3 * AC) )^2 = AC^2
3600 + 1600 + (160/3)AC + (4/9)AC^2 = AC^2
(5/9)AC^2 - (160/3)AC - 5200 = 0
Multiply through by 9
5AC^2 - 480AC - 46800 = 0
AC = [ 480 + sqrt [ 480^2 - 4 * 5 * -46800 ] / 10 =
[ 480 + sqrt [1166400] ] / 10 =
[ 480 + 1080 ] / 10 =
1560 / 10 =
156
