Proof: By Euclid

For, if possible, let them cut one another in half, such that $AE$ is equal to $EC$, and $BE$ to $ED$.
And let the center of the circle $ABCD$ have been found [Prop. 3.1], and let it be (at point) $F$, and let $FE$ have been joined.
Therefore, since some straight line through the center, $FE$, cuts in half some straight line not through the center, $AC$, it also cuts it at right angles [Prop. 3.3].
Thus, $FEA$ is a right angle.
Again, since some straight line $FE$ cuts in half some straight line $BD$, it also cuts it at right angles [Prop. 3.3].
Thus, $FEB$ (is) a right angle.
But $FEA$ was also shown (to be) a right angle.
Thus, $FEA$ (is) equal to $FEB$, the lesser to the greater.
The very thing is impossible.
Thus, $AC$ and $BD$ do not cut one another in half.
Thus, in a circle, if two straight lines, which are not through the center, cut one another then they do not cut one another in half.
(Which is) the very thing it was required to show.
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