# Proof: By Euclid • For, if possible, let it also have been divided at $D$, such that $AD$ and $DB$ are also rational (straight lines which are) commensurable in square only.
• So, (it is) clear that $AC$ is not the same as $DB$.
• For, if possible, let it be (the same).
• So, $AD$ will also be the same as $CB$.
• And as $AC$ will be to $CB$, so $BD$ (will be) to $DA$.
• And $AB$ will (thus) also be divided at $D$ in the same (manner) as the division at $C$.
• The very opposite was assumed.
• Thus, $AC$ is not the same as $DB$.
• So, on account of this, points $C$ and $D$ are not equally far from the point of bisection.
• Thus, by whatever (amount the sum of) the (squares) on $AC$ and $CB$ differs from (the sum of) the (squares) on $AD$ and $DB$, twice the (rectangle contained) by $AD$ and $DB$ also differs from twice the (rectangle contained) by $AC$ and $CB$ by this (same amount) - on account of both (the sum of) the (squares) on $AC$ and $CB$, plus twice the (rectangle contained) by $AC$ and $CB$, and (the sum of) the (squares) on $AD$ and $DB$, plus twice the (rectangle contained) by $AD$ and $DB$, being equal to the (square) on $AB$ [Prop. 2.4].
• But, (the sum of) the (squares) on $AC$ and $CB$ differs from (the sum of) the (squares) on $AD$ and $DB$ by a rational (area).
• For (they are) both rational (areas).
• Thus, twice the (rectangle contained) by $AD$ and $DB$ also differs from twice the (rectangle contained) by $AC$ and $CB$ by a rational (area, despite both) being medial (areas) [Prop. 10.21].
• The very thing is absurd.
• For a medial (area) cannot exceed a medial (area) by a rational (area) [Prop. 10.26].
• Thus, a binomial (straight line) cannot be divided (into its component terms) at different points.
• Thus, (it can be so divided) at one point only.
• (Which is) the very thing it was required to show.

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