1965 AHSME Problem 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 such that CD: DB = 1: 2. The point of intersection of AD and CE is F. Then \frac {EF}{FC} + \frac {AF}{FD} is:

\textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B) }\ \frac {5}{4} \qquad \textbf{(C) }\ \frac {3}{2} \qquad \textbf{(D) }\ 2\qquad \textbf{(E) }\ \frac{5}{2}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.