# $G$ is connected, planar, simple graph with $v$ vertices and $e$ edges. Prove if every face is isomorphic to $C_k$ then $e = \dfrac{k(v-2)}{k-2}$

Let $G$ be a connected, planar, simple graph with $v$ vertices, and $e$ edges.

Prove that if every face is isomorphic to $C_k$ then $e = \dfrac{k(v-2)}{k-2}$

$C_k$ is the cycle of $k$ vertices where $k\geq3$

Here is what I have so far:

Since every face,$f$, is isomorphic to $C_k$, then each face has $k$ vertices and $k$ edges, and every vertex has a degree of $2$.

I'm thinking about using Euler's formula, $v-e+f=2$, but I couldn't fit this formula into my proof.

Any hints about how should I proceed?

Let $\mathcal F$ be the set of faces. Clearly $2E=\sum_{f\in \mathcal F} s(f)$ where $s(f)$ is the number of edges in the face $f$ (this is because every edge is adjacent two exactly two faces and every face is adjacent to $|f|$ edges).
in this particular case we get that $2E=\sum_{f\in \mathcal F} k= kF\iff F=\frac{2E}{k}$.
Now use euler's formula: $V-E+\frac{2E}{k}=2\iff \frac{2-k}{k}E=2-V\iff E=\frac{k(V-2)}{k-2}$