Point E is selected on side AB of triangle ABC in such a way that AE: EB \equal{} 1: 3 and point D is selected on side BC such that CD: DB \equal{} 1: 2. The point of intersection of AD and CE is F. Then \frac {EF}{FC} \plus{} \frac {AF}{FD} is:
<spanclass=′latex−bold′>(A)</span>54<spanclass=′latex−bold′>(B)</span>45<spanclass=′latex−bold′>(C)</span>23<spanclass=′latex−bold′>(D)</span>2<spanclass=′latex−bold′>(E)</span>25