+585 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 $AD$.
+129771 A
10 15
B 1 D 3 C
AD^2 = AB^2 - (BC/4)^2
AD = AC^2 - [(3/4)BC]^2
So
AB^2 - (BC/4)^2 = AC^2 - [(3/4)BC]^2
10^2 - BC^2 / 16 = 15^2 - (9/16)BC^2
15^2 - 10^2 = (9/16 - 1/16)BC^2
125 = (1/2)BC^2
250 = BC^2
AD^2 = 10^2 - (BC/4)^2
AD^2 = 100 - (1/16)BC^2
AD^2 =100 - 250/16
AD^2 = 675 / 8
AD = sqrt (675 / 8) = (15/4)sqrt 6 ≈ 9.19
