Dutch Mathematical Olympiad

a.
Because angles $\angle AEC$ and $\angle ABD$ are straight, we have
$$\angle ABC = 180^{\circ} - \angle DBC = 180^{\circ} - \angle DEC = \angle AED.$$
Because angle $A$ occurs in both triangles, triangles $\triangle ABC$ and $\triangle AED$ have two equal angles, and hence the triangles are similar. ☐

b.
Because of the similarity of triangles $\triangle ABC$ and $\triangle AED$, the angles at $C$ and $D$ are equal. Together with the equality $\angle DBF = \angle CEF$, it follows that triangles $\triangle DBF$ and $\triangle CEF$ are similar.
In a pair of similar triangles, all pairs of sides have the same ratio. Hence, the similarity of triangles $\triangle DBF$ and $\triangle CEF$ yields
$$\frac{|BF|}{|EF|} = \frac{|FD|}{|FC|} = \frac{|DB|}{|CE|} \qquad (1)$$
As triangles $\triangle ABC$ and $\triangle AED$ are similar, we find that
$$\frac{|AB|}{|AE|} = \frac{|BC|}{|ED|} = \frac{|CA|}{|DA|} \qquad (2)$$
Using equations (1) and (2), we can now find $|CF|$. Using the first and last ratio in equation (2), we get $\frac{5}{4} = \frac{|AB|}{|AE|} = \frac{|AC|}{|AD|} = \frac{4+|EC|}{5+3}$. Hence, we have $|EC| = 6$. If we substitute this in the first and third ratio in equation (1), we get $\frac{2}{|EF|} = \frac{3}{6}$. Hence, we have $|EF| = 4$. Using the first and second ratio in (1), we now get that $\frac{2}{4} = \frac{|FD|}{|FC|}$ hence $|FD| = \frac{1}{2}|CF|$. Finally, we substitute this in the first and second ratio in equation (2):
$$\frac{5}{4} = \frac{|AB|}{|AE|} = \frac{|BC|}{|DE|} = \frac{2 + |CF|}{4 + \frac{1}{2}|CF|}.$$
Taking cross ratios and solving the remaining equation, we get $|CF| = 8$. ☐