Perpendicular Bisector of Chord Passes Through Center

Theorem
The perpendicular bisector of any chord of any given circle must pass through the center of that circle.

Proof

 * BisectorOfChord.png

Let $F$ be the center of the circle in question.

Draw any chord $AB$ on the circle.

Bisect $AB$ at $D$.

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

Then the center $F$ lies on $CE$.

The proof is as follows.

Join $FA, FD, FB$.

As $F$ is the center, $FA = FB$.

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

As $FD$ is common, then from Triangle Side-Side-Side Equality, $\triangle ADF = \triangle BDF$.

In particular, $\angle ADF = \angle BDF$; both are right angles.

From :

So $\angle ADF$ and $\angle BDF$ are both right angles.

Thus, by definition, $F$ lies on the perpendicular bisector of $AB$.

Hence the result.

Historical Note
The argument for this particular result originates from Proposition $1$ of Book $\text{III}$ of.

However, the result itself is due to, who reasoned that this result was more fundamental.

This theorem is the converse of.