Proof: By Euclid
(related to Proposition: 3.14: Equal Chords in Circle)
 For let the center of circle $ABDC$^{1} have been found [Prop. 3.1], and let it be (at) $E$.
 And let $EF$ and $EG$ have been drawn from (point) $E$, perpendicular to $AB$ and $CD$ (respectively) [Prop. 1.12].
 And let $AE$ and $EC$ have been joined.
 Therefore, since some straight line, $EF$, through the center (of the circle), cuts some (other) straight line, $AB$, not through the center, at right angles, it also cuts it in half [Prop. 3.3].
 Thus, $AF$ (is) equal to $FB$.
 Thus, $AB$ (is) double $AF$.
 So, for the same (reasons), $CD$ is also double $CG$.
 And $AB$ is equal to $CD$.
 Thus, $AF$ (is) also equal to $CG$.
 And since $AE$ is equal to $EC$, the (square) on $AE$ (is) also equal to the (square) on $EC$.
 But, the (sum of the squares) on $AF$ and $EF$ (is) equal to the (square) on $AE$.
 For the angle at $F$ (is) a right angle [Prop. 1.47].
 And the (sum of the squares) on $EG$ and $GC$ (is) equal to the (square) on $EC$.
 For the angle at $G$ (is) a right angle [Prop. 1.47].
 Thus, the (sum of the squares) on $AF$ and $FE$ is equal to the (sum of the squares) on $CG$ and $GE$, of which the (square) on $AF$ is equal to the (square) on $CG$.
 For $AF$ is equal to $CG$.
 Thus, the remaining (square) on $FE$ is equal to the (remaining square) on $EG$.
 Thus, $EF$ (is) equal to $EG$.
 And straight lines in a circle are said to be equally far from the center when perpendicular (straight lines) which are drawn to them from the center are equal [Def. 3.4] .
 Thus, $AB$ and $CD$ are equally far from the center.
 So, let the straight lines $AB$ and $CD$ be equally far from the center.
 That is to say, let $EF$ be equal to $EG$.
 I say that $AB$ is also equal to $CD$.
 For, with the same construction, we can, similarly, show that $AB$ is double $AF$, and $CD$ (double) $CG$.
 And since $AE$ is equal to $CE$, the (square) on $AE$ is equal to the (square) on $CE$.
 But, the (sum of the squares) on $EF$ and $FA$ is equal to the (square) on $AE$ [Prop. 1.47].
 And the (sum of the squares) on $EG$ and $GC$ (is) equal to the (square) on $CE$ [Prop. 1.47].
 Thus, the (sum of the squares) on $EF$ and $FA$ is equal to the (sum of the squares) on $EG$ and $GC$, of which the (square) on $EF$ is equal to the (square) on $EG$.
 For $EF$ (is) equal to $EG$.
 Thus, the remaining (square) on $AF$ is equal to the (remaining square) on $CG$.
 Thus, $AF$ (is) equal to $CG$.
 And $AB$ is double $AF$, and $CD$ double $CG$.
 Thus, $AB$ (is) equal to $CD$.
 Thus, in a circle, equal straight lines are equally far from the center, and (straight lines) which are equally far from the center are equal to one another.
 (Which is) the very thing it was required to show.^{2}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes