5. Point $A_{1}$ lies on side $B C$ of triangle $A B C$, point $B_{1}$ lies on side $A C$. Let $K$ be the intersection point of segments $A A_{1}$ and $B B_{1}$. Find the area of quadrilateral $A_{1} C B_{1} K$, given that the area of triangle $A_{1} B K$ is 5, the area of triangle $A B_{1} K$ is 8, and the area of triangle $A B K$ is 10.
1.II.5. We will use the following consideration. If a line passes through the vertex of a triangle, then the ratio of the areas of the parts into which it divides the triangle is equal to the ratio of the lengths of the segments into which it divides the opposite side, $\frac{S_{1}}{S_{2}}=\frac{m}{n}$. Let's connect vertex $C$ with point $K$ and set $x=S_{B_{1} C K}$ and $y=S_{A_{1} C K}$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-037.jpg?height=417&width=491&top_left_y=1930&top_left_x=771)

Then

$$
\frac{10}{8}=\frac{B K}{K B_{1}}=\frac{y+5}{x} \text { and } \frac{10}{5}=\frac{A K}{K A_{1}}=\frac{x+8}{y} .
$$

Thus, $5 x=4 y+20$ and $2 y=x+8$. Solving the obtained system, we get that $x=12$ and $y=10$, so $x+y=22$.

Answer: 22.