Proving that togling the direction of a polytope triangle yields the empty set A triangle in the plane can be defined as the intersection of three half planes, each defined by the line passing through one of the segments and a normal direction pointing towards the interior of the triangle.
One thing is, if you flip the direction of all 3 normal directions (i.e. you take the complement of the 3 half spaces) then the intersection of those half spaces is the empty set, no matter the triangle.
I am having a hard time proving that postulate.
 A: In fact, there is a simple reason which I am going to express in a graphical way.
Have a look at the following figure:

The three extended sides (Red, Green, Blue) of the triangle define 7 regions, with a 3 digits binary code for each one.
Let us explain how the first (red) digit is defined.
The regions bearing a left digit number $\color{red}{1}$ are those situated on the same side of the $\color{red}{\text{red}}$ line as the triangle, whereas it is a  $\color{red}{0}$ for the others.
The other digits (the Green and the Blue ones) are attributed with similar rules.
Now the final touch. You may have noticed that among the $2^3=8$ theoretically possible codes, one is absent : code $\color{red}{0}\color{green}{0}\color{blue}{0}$.
What is this missing region, otherwise said, what is the meaning of this void region ? Precisely what you are looking for, the fact that no point can be situated simultaneously at the exterior of red, green and blue lines with respect to the triangle...
A: Simply take any specific point from within that triangle and prove that it will be within neither of those halfspaces. Then observe that this behaviour does not change with the specific choice throughout its inner region.
--- rk
