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 $△ABC$ and $△AED$ have two equal angles, and hence the triangles are similar. ☐

b. Because of the similarity of triangles $△ABC$ and $△AED$, the angles at $C$ and $D$ are equal. Together with the equality $\angle DBF=\angle CEF$, it follows that triangles $△DBF$ and $△CEF$ are similar. In a pair of similar triangles, all pairs of sides have the same ratio. Hence, the similarity of triangles $△DBF$ and $△CEF$ yields $$\frac{|BF|}{|EF|}=\frac{|FD|}{|FC|}=\frac{|DB|}{|CE|}(1)$$ As triangles $△ABC$ and $△AED$ are similar, we find that $$\frac{|AB|}{|AE|}=\frac{|BC|}{|ED|}=\frac{|CA|}{|DA|}(2)$$ 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 $|CF|=8$. ☐