A beautiful proof of the Euler's characteristic of planar graphs goes as follows:
Let $G$ be a connected planar graph and $\chi(G):=V-E+F$. Add as much edges as possible to $G$ (but no vertices!) keeping it's planarity to obtain a maximal planar graph $G'$ so that $G\le G'$. Each added edge divides a face in two, increasing both $E$ and $F$ by $1$, so $\chi(G') = \chi(G)$. Consider a straight line planar embedding of $G'$ (which exists thanks to Fáry's Theorem). Of course every face in $G'$ is a triangle, so each edge belongs to two faces and each face has three edges, therefore $2E = 3F$. Furthermore, each internal face has sum of angles $180^\circ$ and each vertex contributes with $360^\circ$ to these angles, except by the three vertices of the external face, which contributes altogether with $180^\circ$, so $360^\circ(V-3)+180^\circ = 180^\circ(F-1)$ $\therefore 2V-4 = F$. It follows that $\chi(G') = V-E+F = \dfrac{F}{2} + 2 - \dfrac{3F}{2} + F = 2$.
The only reason we require a straight line planar embedding is so that we can double count angles and write an equation relating $V$ and $F$, but is that really necessary? The equation $2V-4 = F$ really doesn't fell like it has anything to do with angles. In fact, because of Fáry's Theorem, we know that this equality is true for any planar embedding of a maximal planar graph.
My issue here is that the proof I know for Fáry's Theorem (which is very elegant) passes through Euler's identity. I know I can prove this identity without Fáry's Theorem by using induction, but I just really love the proof by double counting.
Can we prove that $2V - 4 = F$ for a maximal planar graph without talking about angles nor using induction? Can't we double count something else? Or, at least, using angles without a straight line embedding?