+1234 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 $AD$.
+129895 A
10 15
B 1 D 3 C
Let DC = 3BD
AD = sqrt [ AC^2 - (3BD)^2 ] = sqrt [ 15^2 - 9BD^2 ] (1)
AD = sqrt [ AB^2 - BD^2 ] = sqrt [ 10^2 - BD^2 ] (2)
Equate (1) , (2)
sqrt [ 15^2 - 9BD^2 ] = sqrt [ 10^2 -BD^2 ] square both sides
15^2 - 9BD^2 = 10^2 - BD^2
225 - 9BD^2 = 100 - BD^2
125 = 8BD^2
125/8 = BD^2
AD = sqrt [ 10^2 - BD^2] = sqrt [ 100 - 125/8 ] = sqrt [ 675/8] = [15sqrt 3 ] / [2 sqrt 2] =
7.5 sqrt [3/2] = 7.5 sqrt (6) / 2 = 3.75sqrt (6) ≈ 9.19
