Proof: By Euclid

Let $ADB$ be the given circumference.
So it is required to cut circumference $ADB$ in half.
Let $AB$ have been joined, and let it have been cut in half at (point) $C$ [Prop. 1.10].
And let $CD$ have been drawn from point $C$, at right angles to $AB$ [Prop. 1.11].
And let $AD$, and $DB$ have been joined.

fig30e

And since $AC$ is equal to $CB$, and $CD$ (is) common, the two (straight lines) $AC$, $CD$ are equal to the two (straight lines) $BC$, $CD$ (respectively).
And angle $ACD$ (is) equal to angle $BCD$.
For (they are) each right angles.
Thus, the base $AD$ is equal to the base $DB$ [Prop. 1.4].
And equal straight lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser [Prop. 1.28].
And the circumferences $AD$ and $DB$ are each less than a semicircle.
Thus, circumference $AD$ (is) equal to circumference $DB$.
Thus, the given circumference has been cut in half at point $D$.
(Which is) the very thing it was required to do.¹

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