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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. Therefore:
$$\eqalign{
\frac {DC}{AE} &= \frac{FC}{AF}\cr
\frac{x}{\frac{x}{n}}&= \frac{FC}{AF}\cr
n &= \frac{FC}{AF}
}$$
Hence$$FC = AF \times n $$
So $DE$ cuts $AC$ at the ratio $1:n$.
Folding Fractions
Age 14 to 16
Challenge Level 
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. Therefore:
$$\eqalign{
\frac {DC}{AE} &= \frac{FC}{AF}\cr
\frac{x}{\frac{x}{n}}&= \frac{FC}{AF}\cr
n &= \frac{FC}{AF}
}$$
Hence$$FC = AF \times n $$
So $DE$ cuts $AC$ at the ratio $1:n$.
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