# 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.

In the words of Euclid:

From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line.

## Proof

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.

When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.

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

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

Hence the result.

$\blacksquare$

## Historical Note

The argument for this particular result originates from Proposition $1$ of Book $\text{III}$ of Euclid's The Elements.

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