Question 37
Point $E$ is selected on side $AB$ of triangle $ABC$ in such a way that $AE:EB=1:3,$ and point $D$ is selected on side $BC$ so that $CD:DB=1:2.$ The point of intersection of $AD$ and $CE$ is $F.$ Then
$$
\frac{EF}{FC}+\frac{AF}{FD}
$$
is:
$$
\frac{EF}{FC}+\frac{AF}{FD}
$$
is: