Proof: By Euclid
(related to Proposition: 3.02: Chord Lies Inside its Circle)
 For (if) not then, if possible, let it fall outside (the circle), like $AEB$ (in the figure).
 And let the center of the circle $ABC$ have been found [Prop. 3.1], and let it be (at point) $D$.
 And let $DA$ and $DB$ have been joined, and let $DFE$ have been drawn through.
 Therefore, since $DA$ is equal to $DB$, the angle $DAE$ (is) thus also equal to $DBE$ [Prop. 1.5].
 And since in triangle $DAE$ the one side, $AEB$, has been produced, angle $DEB$ (is) thus greater than $DAE$ [Prop. 1.16].
 And $DAE$ (is) equal to $DBE$ [Prop. 1.5].
 Thus, $DEB$ (is) greater than $DBE$.
 And the greater angle is subtended by the greater side [Prop. 1.19].
 Thus, $DB$ (is) greater than $DE$.
 And $DB$ (is) equal to $DF$.
 Thus, $DF$ (is) greater than $DE$, the lesser than the greater.
 The very thing is impossible.
 Thus, the straight line joining $A$ to $B$ will not fall outside the circle.
 So, similarly, we can show that neither (will it fall) on the circumference itself.
 Thus, (it will fall) inside (the circle).
 Thus, if two points are taken at random on the circumference of a circle then the straight line joining the points will fall inside the circle.
 (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"