# Proving that any planar embedding of $G$ has at least $4$ triangular faces

Let $$G=(V,E)$$ be a connected simple planar graph whose edges can be colored red and blue so that for any vertices $$u,v∈V$$, there is a unique path connecting $$u$$ and $$v$$ whose edges are all red, and a unique path whose edges are all blue.
Prove that any planar embedding of $$G$$ has at least $$4$$ triangular faces (i.e., faces with degree $$3$$). This count may include the outer (exterior)face.

I found that each edge in this graph is contained in a cycle and each face has at least degree $$3$$. Also I have the formula that $$|V| - |E| + |F| = 2$$ and $$3|F| \leqslant 2|E|$$, but I am not sure how to relate this to the number of triangular faces in the graph.

Can you help me solve this problem?

1. Argue that $$E = 2V - 2$$.
2. Argue that if $$G$$ has fewer than $$4$$ triangular faces, then $$2E \geq 4F - 3$$