$_{\text {[the ratio of areas of triangles with a common base or common height] \text { Complexity }}^{\text {Class. }}}$

On side $A B$ of triangle $A B C$, points $M$ and $N$ are taken, such that $A M: M N: N B=2: 2: 1$, and on side $A C$ — point $K$, such that $A K: K C=1: 2$. Find the area of triangle $M N K$, if the area of triangle $A B C$ is 1.
Draw $B K$.

## Solution

Connect $K$ and $B$. Then

$$
S_{\triangle \mathrm{AKB}}=\frac{1}{3} S_{\Delta \mathrm{ABC}}, S_{\Delta \mathrm{MNK}}=\frac{2}{5} S_{\Delta \mathrm{AKB}}
$$

Therefore,

$$
S_{\triangle \mathrm{MNK}}=\frac{2}{5} \cdot \frac{1}{3} S_{\triangle \mathrm{ABC}}=\frac{2}{15} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-29.jpg?height=391&width=912&top_left_y=2194&top_left_x=570)

## Answer

$\frac{2}{15}$.

Point $D$ lies on side $A B$ of triangle $A B C$. Find $C D$, given that $B C=37, A C=15, A B=44, A D$ $=14$.

## Hint

Use the cosine rule to find $\cos \angle A$ from triangle $A B C$.

## Solution

By the cosine rule,

$$
C B^{2}=B A^{2}+A C^{2}-2 B A \cdot B C \cos \angle A
$$

or

$$
37^{2}=44^{2}+15^{2}-2 \cdot 44 \cdot 15 \cdot \cos \angle A
$$

From this, we find that $\cos \angle A=\frac{3}{5}$.

By the cosine rule from triangle $A D C$, we find that

$$
D C^{2}=A D^{2}+A C^{2}-2 A D \cdot A C \cos \angle A=196+225-252=169 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-30.jpg?height=372&width=728&top_left_y=1080&top_left_x=665)

## Answer

13.