C2. We consider a triangle $A B C$ and a point $D$ on the extension of $A B$ towards $B$. The point $E$ lies on the side $A C$ such that angle $\angle D B C$ and angle $\angle D E C$ are equal. The intersection of $D E$ and $B C$ is $F$. Suppose that $|B F|=2,|B D|=3,|A E|=4$ and $|A B|=5$. (Note: the image below is not to scale.)

(a) Show that the triangles $\triangle A B C$ and $\triangle A E D$ are similar.

(b) Determine $|C F|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_88f9acd303931df7c94dg-2.jpg?height=485&width=417&top_left_y=2296&top_left_x=1482)
C2. (a) The situation is as shown in the figure below. Since angles $\angle A E C$ and $\angle A B D$ are straight, it follows that

$$
\angle A B C=180^{\circ}-\angle D B C=180^{\circ}-\angle D E C=\angle A E D .
$$

Since angle $A$ is present in both triangles, triangles $\triangle A B C$ and $\triangle A E D$ have two equal angles and are therefore similar.

(b) Due to the similarity of triangles $\triangle A B C$ and $\triangle A E D$, the angles at $C$ and $D$ are equal. Together with the equality $\angle D B F=\angle C E F$, it follows that triangles $\triangle D B F$ and $\triangle C E F$ are similar.

In similar triangles, all pairs of

![](https://cdn.mathpix.com/cropped/2024_04_17_848643ef5340edd7a266g-4.jpg?height=574&width=439&top_left_y=204&top_left_x=1391)
sides have the same ratio. The similarity of triangles $\triangle D B F$ and $\triangle C E F$ thus gives that

$$
\frac{|B F|}{|E F|}=\frac{|F D|}{|F C|}=\frac{|D B|}{|C E|}
$$

Since triangles $\triangle A B C$ and $\triangle A E D$ are similar, we find the ratios

$$
\frac{|A B|}{|A E|}=\frac{|B C|}{|E D|}=\frac{|C A|}{|D A|}
$$

Using equations (1) and (2), we can now find $|C F|$. From the first and last ratios in equation (2), it follows that

$$
\frac{5}{4}=\frac{|A B|}{|A E|}=\frac{|A C|}{|A D|}=\frac{4+|E C|}{5+3}
$$

From this, it follows that $|E C|=6$. We then substitute this into the first and third ratios in equation (1): $\frac{2}{|E F|}=\frac{3}{6}$. From this, it follows that $|E F|=4$. From the first and second ratios in (1), we now find that $\frac{2}{4}=\frac{|F D|}{|F C|}$, so $|F D|=\frac{1}{2}|C F|$. Finally, we substitute this into the first and second ratios in equation (2):

$$
\frac{5}{4}=\frac{|A B|}{|A E|}=\frac{|B C|}{|D E|}=\frac{2+|C F|}{4+\frac{1}{2}|C F|}
$$

Cross-multiplying and solving the equation gives that $|C F|=8$.