Talei from Poltair Community School and Sports College sent in this excellent solution to the problem, showing the diagonals are split in the ratio 1:2 (or equivalently that the smaller part is a third of the whole)

FD = 2EF and FC = 2AF because . . .

We can prove that triangle FCD is an enlargement, by a scale factor of 2, of the triangle AEF.

The triangles' angles are the same - at point F the crossing lines give each triangle an identical angle because they are opposite each other. The line AC joins parallel lines to give an identical angle to each triangle because they are alternate, and the line ED works in the same way.

Therefore, because AE must be exactly half the length of CD, each line in triangle AEF will be half of its enlarged equivalent in triangle FCD.

FD = 2EF because FD is the enlarged equivalent line of EF in triangle AEF.

FC = 2AF because FC is the enlarged equivalent line of AF in triangle AEF.

Talei's method using similar triangles can also be applied to the general case for a pallelogram and arbitary intersection point:

$$\mathrm{\angle afe = \angle cfd \textrm{ (opposite angles) and } \angle eaf = \angle dcf \textrm{ (alternate / 'Z-' angles)}}$$

$$\mathrm{\therefore \triangle aef \textrm{ is similar to }\triangle cdf \Rightarrow \frac{cf}{af} = \frac{cd}{ae} \Rightarrow cf = \frac{cd}{ae}af}$$

$$\mathrm{ac = af + cf \Rightarrow ac = af \left( 1 + \frac{cd}{ae} \right)}$$

$$\mathrm{\therefore \ af = \frac{1}{1 + \frac{cd}{ae}} \textrm{ multiplying through by ae and using cd = ab} \Rightarrow af = \frac{ae}{ae + ab} ac}$$