In triangle $ABC$, $AB = 10$ and $AC = 15$. Let $D$ be the foot of the perpendicular from $A$ to $BC$. If $BD:CD = 1:3$, then find $AD$.
A
10 15
B D C
AD^2 = 15^2 - 3BD^2
AD^2 = 10^2 - BD^2
AD^2 = AD^2
15^2 - 3BD^2 = 10^2 - BD^2
15^2 -10^2 = 2BD^2
150 = 2BD^2
75 = BD^2
AD^2 = 10^2 - 75
AD^2 = 100 - 75
AD^2 =25
AD = 5
+1790 
