By using Schläfli symbol (namely the fact that $Vq=2E=Fp$) I managed to show that sum of angles of a convex polyhedron is $(E-F)2\pi$, where $E$ - edges, $F$ - faces, $V$ - vertices).
I want to show that the same sum is $(V-2)2\pi$ (*). The problem is I can't use Euler characteristic since it is the final goal of my computation. Also I want to avoid using topology and stay within geometry.
I found two similar proofs one and two but both of them considering polyhedron as a planar graph. So my question is there any approach where I can prove (*) only by using geometry or graph approach is inevitable?
I know that Descartes derived formula $Vq=2F+2V-4$ and from this fact I can proof (*) easily, but the problem I have no clue how to prove his formula.
Another apporach I tried is to do it through angular deficiency but I didn't find a way to show it equals to $4\pi$ without using Euler characteristic.