Condition for Point to be Center of Circle

Proof
Let $ABC$ be a circle, and $D$ a point in it such that $DA = DB = DC$.

Then $D$ can be proved to be the center of circle $ABC$ as follows.


 * Euclid-III-9.png

Join $AB$ and $BC$, and bisect them at $E$ and $F$.

Join $ED$ and $FD$, and produce them in both directions to $GK$ and $HL$.

We have that $AE = EB$, $AD = BD$ and $ED$ is common.

From Triangle Side-Side-Side Equality, $\triangle AED = \triangle BED$ and so $\angle AED = \angle BED$.

So from, each of $\angle AED$ and $\angle BED$ is a right angle.

Hence from the porism to Finding Center of Circle, the center of circle $ABC$ is on $GK$.

The same applies to $\triangle BFD = \triangle CFD$, and so by the same construction, the center of circle $ABC$ is also on $HL$.

The straight lines $GK$ and $HL$ have only point $D$ in common.

Hence the result.