Finding Center of Circle/Proof 1

Proof

 * Euclid-III-1.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.

Suppose $F$ were not the center of the circle, but that $G$ were instead.

Join $GA, GB, GD$.

As $G$ is (as we have supposed) the center, then $GA = GB$.

Also, we have $DA = DB$ as $D$ bisects $AB$.

So from Triangle Side-Side-Side Equality:
 * $\triangle ADG = \triangle BDG$

Hence:
 * $\angle ADG = \angle BDG$

But from :

So $\angle ADG$ is a right angle.

But $\angle ADF$ is also a right angle.

So $\angle ADG = \angle ADF$, and this can happen only if $G$ lies on $CE$.

But then as $G$ is, as we suppose, at the center of the circle, then $GC = GE$.

Thus it follows that $G$ bisects $CE$.

But then $GC = FC$, and so $G = F$.

Hence the result.