+1721 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 $AD$.
+130458 Let BD = 1/4 BC
Let CD = 3/4 BC
So
AB^2 - [(1/4)BC]^2 = AD^2
and
AC^2 - [ (3/4)BC]^2 = AD^2
So....this implies that
AB^2 - [(1/4)BC]^2 = AC^2 -[ (3/4)BC]^2
10^2 - BC^2/16 = 15^2 - (9/16)BC^2
(9/16 - 1/16] BC^2 = 15^2 - 10^2
(15/16] BC^2 = 125
BC ^2 = 125 * 16/15
BC^2 = 400 /3
So
AD^2 = AB^2 - BD^2
AD^2 = AB^2 - BC^2/16
AD^2 = 10^2 - (400/3) /16
AD^2 = 100 - 400/48
AD^2 = 100 - 25/3
AD^2 = [ 300 - 25 ] / 3
AD = 275 / 3
AD = sqrt [ 275 / 3 ] ≈ 9.58
