We can draw a figure like this:
Let us consider the lengths of AB,BC and CA be 2 units, 4 units and 3 units respectively.
As D is the mid point of BC, AD is the median. Also, BD=CD= 2 units.
Using Apollonius theorem,
$AB^2+AC^2=2\left(AD^2+BD^2\right)$
or,$2^2+3^2=2\left(AD^2+2^2\right)$
or,$AD^2=\dfrac{5}{2}$
or,$AD=\dfrac{\sqrt{\ 10}}{2}$ ------(1)
Now,$AM$ is perpendicular to $BC$.
So,$\triangle\ AMB$ and$\triangle\ AMC$ both are right angled triangles.
Using Pythagoras theorem in$\triangle\ AMB$, we can write,
$AM^2=AB^2-BM^2$
or,$AM^2=2^2-BM^2$ -----(2)
Similarly,using Pythagoras theorem in $\triangle\ AMC$, we can write,
$AM^2=AC^2-CM^2$
or,$AM^2=3^2-\left(4-BM\right)^2$ -------(3)
From equation (2) and (3),
$3^2-\left(4-BM\right)^2=2^2-BM^2$
or,$3^2-2^2=\left(4-BM\right)^2-BM^2$
or,$5=\left(4-2BM\right)\cdot4$
or,$BM=\dfrac{11}{8}$ -----(4)
From equation (1) and (4),
$\dfrac{BM}{AD}=\dfrac{\dfrac{11}{8}}{\dfrac{\sqrt{\ 10}}{2}}=\dfrac{11}{4\sqrt{\ 10}}$
So,$BM:AD=11:4\sqrt{\ 2}$
