4. In triangle $A B C$, side $A C$ is greater than $A B$, line $l$ is the bisector of the external angle $C$. A line passing through the midpoint of $A B$ and parallel to $l$ intersects $A C$ at point $E$. Find $C E$, if $A C=7$ and $C B=4$. (The external angle of a triangle is the angle adjacent to the internal angle at the given vertex.)

$(25$ points. $)$
Answer: $\frac{11}{2}$.

(Fig. 1.) Let $F$ be the intersection point of line $l$ and line $A B$. Since $A C>A B$, point $E$ lies on segment $A C$, and point $F$ lies on ray $A B$. Draw a line through point $B$ parallel to $A C$ until it intersects $C F$ at point $G$. Then $\angle B G C=\angle B C G$, so triangle $C B G$ is isosceles, $B G=B C$. From the similarity of triangles $A F C$ and $B F G$, we have:

$$
\frac{F A}{F B}=\frac{A C}{B G}=\frac{A C}{B C}=\frac{7}{4} \quad \Leftrightarrow \quad \frac{A B}{A F}=\frac{3}{7} \quad \Leftrightarrow \quad \frac{A O}{A F}=\frac{A B / 2}{A F}=\frac{3}{14}
$$

Since triangles $A C F$ and $A E O$ are also similar, we have:

$$
\frac{A E}{A C}=\frac{A O}{A F}=\frac{3}{14}
$$

Therefore, $A E=\frac{3}{14} A C=\frac{3}{2}$ and $E C=A C-A E=\frac{11}{2}$.

Criteria. The similarity of triangles $A F C$ and $B F G$ is proven - 10 points, the similarity of $A C F$ and $A E O$ is proven - another 10 points.

![](https://cdn.mathpix.com/cropped/2024_05_06_b9c9fe6aa9b1b5d44cacg-2.jpg?height=360&width=873&top_left_y=1522&top_left_x=664)

Fig. 1