Regular Polygon is Cyclic

Theorem

Let $P$ be a regular polygon.

Then $P$ is a cyclic polygon.

Proof

Let $P$ be a regular polygon.

Aiming for a contradiction, suppose $P$ is not cyclic.

Let $AB$, $BC$ and $CD$ be sides of $P$ such that $A$, $B$ and $C$ are on the circumference of a circle $K$ such that $D$ is not on that circumference.

This has to be possible, or all vertices of $P$ would lie on $K$.

That would make $P$ cyclic which contradicts our supposition about $P$.

Let the center of $K$ be $O$.

As $D$ is not on the circumference of that circle, then $OC \ne OD$.

Hence $\angle ABC \ne \angle BCD$ and so $P$ is not equiangular.

Hence a fortiori $P$ is not a regular polygon.

This contradicts our assertion about $P$.

Hence by Proof by Contradiction it follows that $P$ is cyclic.

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): polygon
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): polygon