[Example 1.4.5] On the plane, there are finitely many polygons. If any two of these polygons are intersected by a line passing through the origin, then these polygons are said to be properly placed. Find the smallest natural number $m$, such that for any set of properly placed polygons, one can draw $m$ lines passing through the origin, so that each of these polygons intersects at least one of these $m$ lines.
To facilitate the study of the problem, we introduce the concept of the angle subtended by a polygon at point $O$.
Drawing rays from point $O$ to every point of polygon $M$, there must be a largest angle among these rays, and this largest angle is called the angle subtended by polygon $M$ at point $O$.

As shown in Figure 1-24, $\angle AOB$ is the angle subtended by polygon $M$ at point $O$. Clearly, every point of polygon $M$ lies within the region of $\angle AOB$. To find the minimum value of $m$, the polygon with the smallest angle subtended at $O$ plays a crucial role.
Let's start with simple cases.
If there are only two polygons properly placed, by the definition of proper placement, $m=1$.

If there are three polygons, and they are properly placed, as shown in Figure 1-25, polygons $M_1, M_2, M_3$ are properly placed. From the figure, it can be seen that $M_1$ and $M_3$ must have a line passing through the origin that intersects them, and the same applies to $M_2$ and $M_3$, while $M_1$ and $M_2$ are intersected by the $y$-axis.
Clearly, $m=1$ does not meet the requirement, so $m \geqslant 2$.
Now consider a finite number of polygons that are properly placed.

For this finite number of polygons, there is an extreme case where one polygon has the smallest angle subtended at $O$. Let polygon $M$ have the smallest angle $\alpha$ subtended at $O$, with the lines on which the sides of $\alpha$ lie being $l_1$ and $l_2$.

We study the relationship between other polygons and these two lines $l_1$ and $l_2$.

Let polygon $M'$ be any polygon different from $M$.

By the problem's conditions, there must be a line passing through $O$ that intersects both polygon $M$ and polygon $M'$. Let this line be $l$.
Clearly, $l$ either coincides with one of $l_1$ and $l_2$, or lies between $l_1$ and $l_2$.
If $l$ coincides with one of $l_1$ and $l_2$, then $m=2$.
If $l$ lies between $l_1$ and $l_2$, then since the angle $\beta$ subtended by $M'$ at the origin is $\beta \geqslant \alpha$, there must be a line $l'$ passing through the origin that intersects $M'$ and lies outside $\angle \alpha$, or coincides with one of $l_1$ and $l_2$.
If $l'$ coincides with one of $l_1$ and $l_2$, then $m=2$.
If $l'$ lies outside $\angle \alpha$, then one of $l_1$ and $l_2$ must lie within the angle $\beta$ between $l$ and $l'$. Without loss of generality, let it be $l_1$, so polygon $M'$ intersects with line $l_1$.

The above proof shows that polygon $M'$ intersects with at least one of the two lines $l_1$ and $l_2$.
Therefore, the smallest natural number $m=2$.