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.

(The Elements: Book $\text{III}$: Proposition $1$ : Porism)

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 Congruence, $\triangle ADF = \triangle BDF$.

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

From Book $\text{I}$ Definition $10$: Right Angle:

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.

This theorem is the converse of Proposition $3$ of Book $\text{III} $: Conditions for Diameter to be Perpendicular Bisector.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions