Finding Center of Circle/Proof 2

Proof

 * Euclid-III-1-Proof-2.png

Draw any chord $AB$ on the circle in question.

Bisect $AB$ at $D$.

Construct $CE$ perpendicular to $AB$ at $D$, where $C$ and $E$ are where this perpendicular meets the circle.

Bisect $CE$ at $F$.

Then $F$ is the center of the circle.

The proof is as follows.

From Perpendicular Bisector of Chord Passes Through Center, $CE$ passes through the center of the circle.

The center must be the point $F$ such that $FE = FC$.

That is, $F$ is the bisector of $CE$.

Historical Note
This proof was formulated by who preferred to prove the more fundamental result first, wording it as:
 * The line which bisects a chord perpendicularly must contain the center

and then use that to prove this.