Conditions for Diameter to be Perpendicular Bisector

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If in a circle a diameter bisects a chord (which is itself not a diameter), then it cuts it at right angles, and if it cuts it at right angles then it bisects it.

In the words of Euclid:

If in a circle a straight line through the centre bisect a straight line not through the centre, then it cuts it at right angles; and if it cuts it at right angles, it also bisects it.

(The Elements: Book $\text{III}$: Proposition $3$)



Let $ABC$ be a circle, in which $AB$ is a chord which is not a diameter (i.e. it does not pass through the center).

First part

Let $CD$ be a diameter which bisects $AB$ at the point $F$.

Find the center $E$ of circle $ABC$, and join $EA$ and $EB$.

Because $E$ is the center, $EA = EB$.

Because $F$ bisects $AB$, $FA = FB$.

Also, $FE$ is common, so from Triangle Side-Side-Side Equality $\triangle AFE = \triangle BFE$.

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 $\angle AFE$ and $\angle BFE$ are both right angles.

Therefore the diameter $CD$ cuts $AB$ at right angles.


Second part

Let $CD$ be a diameter which cuts $AB$ at right angles at point $F$.

We using the same construction as above.

Because $E$ is the center, $EA = EB$.

From Isosceles Triangle has Two Equal Angles, $\angle EAF = \angle EBF$.

But $\angle AFE = \angle BFE$ because both are right angles.

As $EF$ is common, it follows from Triangle Side-Angle-Angle Equality that $\triangle AFE = \triangle BFE$.

Therefore $AF = FB$.

Hence the result.


Historical Note

This proof is Proposition $3$ of Book $\text{III}$ of Euclid's The Elements.
It is the converse of the Porism to Proposition $1$: Perpendicular Bisector of Chord Passes Through Center.