## Task B-1.5.

The area of triangle $A B C$ is $32 \mathrm{~cm}^{2}$. Points $M$ and $N$ are chosen on sides $A B$ and $B C$ such that $|A M|:|M B|=|B N|:|N C|=3: 1$. Point $G$ is the intersection of lines $A N$ and $C M$. Determine the ratio of the areas of figures $A G C$ and $M B N G$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-03.jpg?height=565&width=511&top_left_y=1342&top_left_x=658)

Let's determine the areas of triangles $A N C$ and $B C M$.

$$
P_{A N C}=\frac{1}{2} v_{a} \cdot \frac{1}{4} a=\frac{1}{4} \cdot \frac{1}{2} a v_{a}=\frac{1}{4} P_{A B C}=\frac{1}{4} \cdot 32=8 \mathrm{~cm}^{2}
$$

Similarly, we get

$$
P_{A M B}=\frac{1}{2} v_{c} \cdot \frac{1}{4} c=\frac{1}{4} P_{A B C}=8 \mathrm{~cm}^{2}
$$

If triangles $A N C$ and $A M B$ have the same area, then triangles $A G C$ and quadrilateral $B N G M$ have the same area, so the desired ratio is $1: 1$.