So I'm working on generalizing the Saccheri-Legendre Theorem to convex $n$-gons.
$\underline{\text{Saccheri-Legendre Theorem:}}$ The sum of the angles of a triangle is at most $180^\circ$.
A triangle is an $n$-gon where $n=3$. So to generalize I would want a formula for any $n$. Which follows, I believe, the same formula as the formula in Euclidean geometry for angle measure of a triangle. So I could say the following.
$\underline{\text{Generalization of the Sacceri-Legendre Theorem:}}$ For any $n$-gon the angle measure of that $n$-gon is at most $2\cdot(n-2)\cdot90^\circ$.
So for my proof I'm assuming that I'm going to have to show that any $n$-gon can break up into $m$ different triangles. Then I can say by the Saccheri-Legendre Theorem that all of the $m$ different triangles add up to $180^\circ$. That way it would just be $m$ different Saccheri-Legendre proofs which I can easily prove since we've used it before.
Just want to make sure that I am on the right track and not looking at this completely wrong.
You're right. Every convex $n$-gon can be divided into $n-2$ triangles. So the sum $S$ of the internal angles of the $n$-gon is $(n-2)S_\triangle \leq (n-2)180^\circ$, where $S_\triangle$ is the sum of the internal angles of the triangle (if you want to apply Saccheri-Legendre for triangles right of the bat).