Why Does a Polygon Have the Same Number of Vertices and Sides?

For an ordinary polygon, the number of vertices equals the number of sides because its boundary is a closed chain: each side connects one vertex to the next, and the final side returns to the first vertex. The same count can also be proved by induction.

The direct counting argument

Suppose a polygon has vertices $v_1,v_2,\ldots,v_n$, listed in order around its boundary. Its sides are

$$v_1v_2,\ v_2v_3,\ \ldots,\ v_{n-1}v_n,\ v_nv_1.$$

There are $n-1$ sides joining successive vertices in the list, plus one closing side from $v_n$ to $v_1$. Therefore the total number of sides is

$$(n-1)+1=n.$$

So the polygon has $n$ vertices and $n$ sides. Equivalently, each vertex has one outgoing boundary side, and following these sides once around the boundary returns you to your starting point.

An inductive proof

You can make the count explicit with induction. First consider an open chain of vertices, before adding the closing side.

Base case: A chain consisting of one vertex has no connecting sides. Thus, for one vertex, the number of sides in the open chain is $0=1-1$.

Inductive step: Suppose an open chain of $k$ vertices has $k-1$ sides. Add one more vertex at the end. It creates exactly one new side, connecting the old last vertex to the new one. The chain now has $k+1$ vertices and

$$(k-1)+1=k$$

sides. Thus an open chain of $n$ vertices has $n-1$ sides.

To turn that chain into a polygon, add the closing side from the last vertex back to the first. The polygon then has $(n-1)+1=n$ sides. It has the same $n$ vertices, as required. A polygon has at least three vertices, but the chain-counting argument works for any number of vertices.

Example: a pentagon

List the five vertices as $A,B,C,D,E$ in boundary order. The sides are $AB$, $BC$, $CD$, $DE$, and $EA$. There are five vertices and five sides. The last side, $EA$, matters: without it, the list describes an open chain rather than a polygon.

The argument does not depend on the polygon being regular, convex, or having equal side lengths. It depends only on the boundary being a closed sequence of sides and vertices.

Common mistakes and conventions

• Forgetting the closing side: A list of $n$ vertices has only $n-1$ links if you stop at the last listed vertex. A polygon includes the link from the last vertex back to the first.
• Confusing sides with side lengths: The claim concerns the number of side segments, not their lengths. A polygon’s sides can have different lengths.
• Counting collinear points differently: Some definitions allow a marked point along a straight side to count as a vertex, while others reserve “vertex” for a corner where the direction changes. Under the usual polygon convention, the counted sides are the boundary segments between consecutive counted vertices. If a straight side is split by an extra marked point, count that point as a vertex and the two resulting segments as sides; the equality still holds.
• Using the claim for an open chain: An open chain with $n$ vertices has $n-1$ connecting segments. The extra segment that closes the chain is exactly why a polygon has $n$ sides.

The essential idea is cyclic order: once you decide which boundary points are the vertices, each consecutive pair—including the last and first—defines one side.