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.

Geometric Proof

 * BisectorOfChord.png

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

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 book 1 definition 10:
 * "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.

Note
The argument for this particular result originates from Proposition 1 of Book III of Euclid's.

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 III of Euclid's : Conditions for Diameter to be Perpendicular Bisector.