Proof: By Euclid
(related to Proposition: 3.03: Conditions for Diameter to be a Perpendicular Bisector)
 For let the center of the circle $ABC$ have been found [Prop. 3.1], and let it be (at point) $E$, and let $EA$ and $EB$ have been joined.
 And since $AF$ is equal to $FB$, and $FE$ (is) common, two (sides of triangle $AFE$) [are] equal to two (sides of triangle $BFE$).
 And the base $EA$ (is) equal to the base $EB$.
 Thus, angle $AFE$ is equal to angle $BFE$ [Prop. 1.8].
 And when a straight line stood upon (another) straight line makes adjacent angles (which are) equal to one another, each of the equal angles is a right angle [Def. 1.10] .
 Thus, $AFE$ and $BFE$ are each right angles.
 Thus, the (straight line) $CD$, which is through the center and cuts in half the (straight line) $AB$, which is not through the center, also cuts ($AB$) at right angles.
 And so let $CD$ cut $AB$ at right angles.
 I say that it also cuts ($AB$) in half.
 That is to say, that $AF$ is equal to $FB$.
 For, with the same construction, since $EA$ is equal to $EB$, angle $EAF$ is also equal to $EBF$ [Prop. 1.5].
 And the right angle $AFE$ is also equal to the right angle $BFE$.
 Thus, $EAF$ and $EFB$ are two triangles having two angles equal to two angles, and one side equal to one side  (namely), their common (side) $EF$, subtending one of the equal angles.
 Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
 Thus, $AF$ (is) equal to $FB$.
 Thus, in a circle, if any straight line through the center cuts in half any straight line not through the center then it also cuts it at right angles.
 And (conversely) if it cuts it at right angles then it also cuts it in half.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"