In triangle $ABC$, the side lengths are $AB = BC = 5$ and $AC = 6.$ If $AD$ is an angle bisector such that $D$ is on $BC,$ then find the value of $AD^2.$ Express your answer as a common fraction.
+129771 
Angle Bisector Theorem
BD /AB = DC/ AC let BD = x ......so DC = BC - x = 5 - x
x/ 5 = (5-x) / 6
6x = 5(5-x)
6x = 25 - 5x
11x = 25
x = 25/11 = BD
Law of Cosines
AC^2 = AB^2 + BC^2 - 2(AB * BC) cos ABC
6^2 = 5^2 + 5^2 - 2(5 * 5)cos ABC
[36 - 50 ] / [ -50] = cos ABC
cosABC = -14/-50 = 7/25
Law of Cosines again
AD^2 = AB^2 + BD^2 - 2(AB * BC) cos ABC
AD^2 = 5^2 + (5/11)^2 - 2 ( 5 * 5/11) (7/25)
AD^2 = 25 + 25/121 - (50/11)(7/25)
AD^2 = 2896 / 121
