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