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If a finite simple graph is a connected graph, a bipartite graph, and a planar graph, with $n\ge 3$ vertices, then $m\le 2n-4$. Every facial boundary walk has even length and length at least three, hence at least four. Thus $4f\le 2m$; substitute the Euler formula for a connected planar graph.^[ref]

The square cycle graph attains equality: $4=2\cdot 4-4$. The three-vertex graph path also attains equality: $2=2\cdot 3-4$. Its repeated bridges in a graph illustrate why counting facial boundary walk length correctly matters. Neither bound applies unchanged to a multigraph: two vertices may support arbitrarily many parallel edges in a plane graph.

For a finite connected graph with at least one vertex, embedded as a plane graph, let $n$ count vertices, $m$ count edges, and $f$ count faces of a plane graph, including the outer face. ThenThe hypotheses refer to an actual embedding; regions in a crossing-filled graph drawing do not supply its $f$.^[ref]

In the plane graph $K_4$ pictured above, three bounded faces of a plane graph and the outer face give $4-6+4=2$. A square cycle graph gives $4-4+2=2$, while the graph path $A-B-C$ gives $3-2+1=2$. Changing a crossing into a new vertex changes the graph: the square-and-diagonals drawing then has five vertices, eight edges, and five faces of a plane graph, giving $5-8+5=2$.

A proof outline deletes edges on graph cycles. Each deletion preserves the connected graph and merges two faces of a plane graph, leaving $n-m+f$ unchanged. Eventually a tree remains, with $m=n-1$ and $f=1$. Adding the point at infinity identifies the plane picture with a sphere; the value two agrees with its Euler characteristic and the cell-count formula for Euler characteristic.

For a finite connected graph embedded as a plane graph, follow the edges around a face of a plane graph to obtain a closed walk. Count traversals, not distinct edges: a bridge in a graph occurs twice along one facial boundary walk. Each edge contributes two sides in total.^[ref]

For the three-vertex graph path $A-B-C$, the only facial boundary walk is $A,B,C,B,A$, of length four. Both edges are bridges in a graph; neither is a graph cycle. Thus a facial boundary walk need not be a graph cycle.

The outer face is the unique unbounded face of a plane graph for a finite plane graph. It counts once even though it surrounds the whole graph drawing. A tree has only this face of a plane graph.

For a finite connected plane graph, with $v$ vertices, $e$ edges and $f$ faces including the unbounded face, Euler formula for a connected planar graph isIf the graph is a tree, $e=v-1$ and $f=1$. Otherwise remove an edge on a cycle. This preserves connectivity and joins the two faces on its sides, so both $e$ and $f$ decrease by one. Repeating reaches a tree, preserving $v-e+f$ throughout and proving the formula.

For a simple connected graph with $v\ge 3$, each face boundary walk has length at least three, with bridges counted twice in the walk; the sum of lengths is $2e$. Hence $3f\le 2e$, which with Euler gives $e\le 3v-6$. If it has no triangles in a graph, no face boundary walk can have length three, and every boundary length is at least four. Thus $4f\le 2e$, givingFor a disconnected graph, add noncrossing bridges between its components; this preserves simplicity and creates no triangles, reducing to the connected bound.

Every triangle-free planar graph with at least three vertices has average degree $2e/v<4$, and therefore a vertex of degree at most three. Remove one such vertex. Its remaining graph is again triangle-free and planar. Induction, with graphs on at most two vertices as the base cases, provides a graph colouring with four colours of the remainder. At most three colours appear on the removed vertex's neighbours, so a fourth colour extends the graph colouring. This proves four-colourability without using the four-colour theorem.

A planar graph admits at least one graph drawing in which edges meet only at common endpoints. This is a property of the abstract graph, not a requirement on every graph drawing. A plane graph specifies one such drawing.^[ref]

For example, the complete graph $K_4$ joins every pair of four vertices, giving six edges. Put its vertices at the corners of a square and its diagonals cross. Instead put $A,B,C$ at the corners of a triangle and $D$ inside it: the same six edges $AB,AC,BC,AD,BD,CD$ now meet only at their endpoints. The first graph drawing obscures planarity; the second proves it. In particular, its crossing number is zero.

The distinction matters whenever connections must fit on one surface without intersecting. A failed graph drawing may only need rearrangement, whereas nonplanarity by edge counts proves that every rearrangement fails.

A finite simple graph that is a connected graph and a planar graph, with $n\ge 3$ vertices and $m$ edges, satisfies $m\le 3n-6$. Choose a plane graph: each facial boundary walk has length at least three, so $3f\le 2m$. Substitute $f=2-n+m$ from the Euler formula for a connected planar graph.^[ref]

The complete graph $K_4$ attains equality: $6=3\cdot 4-6$. The bound also holds for a disconnected finite simple graph that is a planar graph with $n\ge 3$: place its connected components of a graph in disjoint regions and join them with new bridges in a graph, then apply the connected bound.

A plane graph is a planar graph with a specified graph drawing whose edges meet only at common endpoints, also called an embedding in the Euclidean plane. Its faces of a plane graph and facial boundary walks refer to that embedding. This geometric use of embedding differs from a graph embedding, which maps one abstract graph into another.