Proof: By Euclid
(related to Proposition: 3.14: Equal Chords in Circle)
- For let the center of circle $ABDC$ 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.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes
