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# 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 $1:n$.