In △ A B C, side BC is divided by D in a ratio of...

Step 1
A general method to solving these kinds of things is through the Ratio Lemma, which is often applied in high school olympiad geometry. It states that if AD is a cevian in $\mathrm{△}ABC$, then ${\displaystyle \frac{BD}{DC}}={\displaystyle \frac{AB}{AC}}\cdot {\displaystyle \frac{\sin \mathrm{\angle }BAD}{\sin \mathrm{\angle }CAD}}$. The proof of this is quite simple; just apply the sine law to triangles ABD and CAD.
Step 2
So for this problem, the ratios $\sin \mathrm{\angle }BAD:\sin \mathrm{\angle }CAD$ is clearly equal to $\sin \mathrm{\angle }EAF:\sin \mathrm{\angle }CAF$. So $EF:FC$ is equal to $(BD/DC)(AE/AB)=10/7$. $DF:FA$ is calculated similarly.
The Ratio Lemma can also be used to prove many basic facts in projective geometry; e.g. the invariance of the cross-ratio when projected through a point onto a line.

Step 1
Use mass points. We assign B a mass of 4. So A has a mass of 3 and C has a mass of 10. E has a mass of 7 and D has a mass of 14.
Step 2
So we have $\frac{AF}{FD}=\frac{14}{3}$
then $\frac{EF}{FC}=\frac{10}{7}$