If AE/DE = 3, then what is AD/BC?
(Also, B is the midpoint of AE, and C is the midpoint of AD.)
Triangle ACB is similar to triangle ADE
Let AE = 6
AE / DE = 3
Then DE = 2
Then AD = sqrt (6^2 + 2^2) = sqrt 40 = 2sqrt 10
So AC = 1/2 AD = sqrt 10
And AB = (1/2) AE = 3
So
BC = sqrt [ (sqrt (10)^2 - 3^2 ] = sqrt 1 = 1
So
AD / BC = 2 sqrt ( 10 ) / 1 = 2 sqrt 10
EDIT to correct a previous error !!!
Let \(DE = x\)
We know that \(AE = 3x\) and \(AD = \sqrt{10}x\)
We also know that \(\triangle ABC\) is similar to \(\triangle ADE\)
Because the scale factor is 2, we know that \(CB = 0.5x\)
Thus, the ratio is \({\sqrt{10} \over 0.5 } = \color{brown}\boxed{2\sqrt{10}}\)