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.

Proof

 * Euclid-III-4.png

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 $E$.

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 intersect each other.