(related to Proposition: 3.35: Intersecting Chord Theorem)

- In fact, if $AC$ and $BD$ are through the center (as in the first diagram from the left), so that $E$ is the center of circle $ABCD$, then (it is) clear that, $AE$, $EC$, $DE$, and $EB$ being equal, the rectangle contained by $AE$ and $EC$ is also equal to the rectangle contained by $DE$ and $EB$.
- So let $AC$ and $DB$ not be though the center (as in the second diagram from the left), and let the center of $ABCD$ have been found [Prop. 3.1], and let it be (at) $F$.
- And let $FG$ and $FH$ have been drawn from $F$, perpendicular to the straight lines $AC$ and $DB$ (respectively) [Prop. 1.12].
- And let $FB$, $FC$, and $FE$ have been joined.
- And since some straight line, $GF$, through the center, cuts at right angles some (other) straight line, $AC$, not through the center, then it also cuts it in half [Prop. 3.3].
- Thus, $AG$ (is) equal to $GC$.
- Therefore, since the straight line $AC$ is cut equally at $G$, and unequally at $E$, the rectangle contained by $AE$ and $EC$ plus the square on $EG$ is thus equal to the (square) on $GC$ [Prop. 2.5].
- Let the (square) on $GF$ have been added [to both].
- Thus, the (rectangle contained) by $AE$ and $EC$ plus the (sum of the squares) on $GE$ and $GF$ is equal to the (sum of the squares) on $CG$ and $GF$.
- But, the (square) on $FE$ is equal to the (sum of the squares) on $EG$ and $GF$ [Prop. 1.47], and the (square) on $FC$ is equal to the (sum of the squares) on $CG$ and $GF$ [Prop. 1.47].
- Thus, the (rectangle contained) by $AE$ and $EC$ plus the (square) on $FE$ is equal to the (square) on $FC$.
- And $FC$ (is) equal to $FB$.
- Thus, the (rectangle contained) by $AE$ and $EC$ plus the (square) on $FE$ is equal to the (square) on $FB$.
- So, for the same (reasons), the (rectangle contained) by $DE$ and $EB$ plus the (square) on $FE$ is equal to the (square) on $FB$.
- And the (rectangle contained) by $AE$ and $EC$ plus the (square) on $FE$ was also shown (to be) equal to the (square) on $FB$.
- Thus, the (rectangle contained) by $AE$ and $EC$ plus the (square) on $FE$ is equal to the (rectangle contained) by $DE$ and $EB$ plus the (square) on $FE$.
- Let the (square) on $FE$ have been taken from both.
- Thus, the remaining rectangle contained by $AE$ and $EC$ is equal to the rectangle contained by $DE$ and $EB$.
- Thus, if two straight lines in a circle cut one another then the rectangle contained by the pieces of one is equal to the rectangle contained by the pieces of the other.
- (Which is) the very thing it was required to show.
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∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"