Source:
Mathematical Circles
The solution to this problem centers around how convex $n$-gons cannot have more than two mutually parallel sides. In other words, each side of a convex $n$-gon can be parallel to another side but no more. For example, a square can have $2$ sets of $2$ mutually parallel sides but we can't have $1$ set of $3$ mutually parallel sides.
Here is my informal attempt at proving that a convex $n$-gon can have at most two mutually parallel sides.
Pick a side of an $n$-gon and we'll use that as a reference line. In order to get the angle of any line with reference to the current reference line, it is sufficient to consider the sum of the exterior angles from this reference line up to the desired line. See Figure 1 below.
In the figure, the angle the side with $C^{\circ}$ makes against the reference base line with the $A^{\circ}$ is equal to $A^{\circ}+B^{\circ}+C^{\circ}$. So for a line to be parallel to the base line, the angle between this line and the base line must be a multiple of $180^\circ$ since $\sin(n*180^\circ) = 0$ for all $n\in\mathbb{N}$.
Note that a line having an angle of $0^\circ$ or $360^\circ$ to a reference line is the exact same line as the reference line. Then since the sum of exterior angles of a convex $n$-gon equals $360^{\circ}$, there is only one other option for a mutually parallel line and so a convex $n$-gon can only have at most $2$ mutually parallel lines.
Questions:
1) What's the standard way of proving this?
2) What's the quickest way of proving this?