Richard of Mearns Castle High School sent us in this solution:

Notice that $AE = \frac{1}{n} AB = \frac{1}{n}x$

We can say that $\triangle AEF$ and $\triangle DCF$ are similar. This is because $\angle AFE$ is equal to $\angle DFC$ (vertically opposite). Also $\angle FAE = \angle FCD$ and $\angle AEF = \angle CDF$ (alternate angles).

Since the triangles are similar we can say that the ratios of corresponding sides are the same. Therfore:

$$\eqalign{

\frac {DC}{AE} &= \frac{FC}{AF}\cr

\frac{x}{\frac{x}{n}}&= \frac{FC}{AF}\cr

n &= \frac{FC}{AF}

}$$

Hence$$FC = AF \times n $$

So $DE$ cuts $AC$ at the ratio $1:n$.