# Chords do not Bisect Each Other

## Theorem

If in a circle two chords (which are not diameters) cut one another, then they do not bisect one another.

In the words of Euclid:

*If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another.*

(*The Elements*: Book $\text{III}$: Proposition $4$)

## Proof

Let $ABCD$ be a circle, in which $AC$ and $BD$ are chords which are not diameters (i.e. they do not pass through the center).

Let $AC$ and $BD$ intersect at $E$.

Aiming for a contradiction, suppose they were able to bisect one another, such that $AE = EC$ and $BE = ED$.

Find the center $F$ of the circle, and join $FE$.

From Conditions for Diameter to be Perpendicular Bisector, as $FE$ bisects $AC$, then it cuts it at right angles.

So $\angle FEA$ is a right angle.

Similarly, $\angle FEB$ is a right angle.

So $\angle FEA = \angle FEB$, and they are clearly unequal.

From this contradiction, it follows that $AC$ and $BD$ can not bisect each other.

$\blacksquare$

## Historical Note

This proof is Proposition $4$ of Book $\text{III}$ of Euclid's *The Elements*.

## Sources

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