A geometry problem by Prem Chebrolu

Consider $\triangle ABC$ . By the angle bisector theorem , we have

$\dfrac{BD}{DC}=\dfrac{AB}{BD}$

$BD=\dfrac{3(2)}{4}=\dfrac{6}{4}=\dfrac{3}{2}$

Consider $\triangle ABD$ . Again, by the angle bisector theorem , we have

$\dfrac{AM}{MD}=\dfrac{AB}{BD}=\dfrac{2}{\dfrac{3}{2}}=\dfrac{4}{3}$

The required answer is

$AM:MD \times 30 = \dfrac{4}{3} \times 30 = 4 \times 10=\boxed{40}$