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.

Geometric 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.