Let $S_{\triangle A B C}=S$. Then

$$
\frac{S_{\Delta \mathrm{H} M K}}{S}=\frac{A M}{A B} \cdot \frac{A K}{A C}=\frac{1}{2} \cdot \frac{A K}{A C}
$$

Since according to the problem, $\frac{S_{\Delta+\Lambda M K}}{S}=\frac{25}{75}=\frac{1}{3}$, we have $\frac{1}{2} \cdot \frac{A K}{A C}=\frac{1}{3}$, from which we find that $\frac{A K}{A C}=\frac{2}{3}$, and $\frac{C K}{A C}=\frac{1}{3}$. Similarly, we find that $\frac{C N}{B C}=\frac{3}{5}$. Therefore,

$$
S_{\triangle C N K}=\frac{C K}{A C} \cdot \frac{C N}{B C} \cdot S=\frac{1}{3} \cdot \frac{3}{5} \cdot 75=15
$$

## Answer

15.
[ Ratio of areas of triangles with a common angle ]

On the sides $A B, B C$ and $A C$ of triangle $A B C$, whose area is 75, points $M, N$ and $K$ are located respectively. It is known that $M$ is the midpoint of $A B$, the area of triangle $B M N$ is 15, and the area of triangle $A M K$ is 25. Find the area of triangle $C N K$.

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