Proof: By Euclid
(related to Proposition: Prop. 10.031: Constructing Medial Commensurable in Square I)
- Let two rational (straight lines), $A$ and $B$, commensurable in square only, be laid out, such that the square on the greater $A$ is larger than the (square on the) lesser $B$ by the (square) on (some straight line) commensurable in length with ($A$) [Prop. 10.29].
- And let the (square) on $C$ be equal to the (rectangle contained) by $A$ and $B$.
- And the (rectangle contained by) $A$ and $B$ (is) medial [Prop. 10.21].
- Thus, the (square) on $C$ (is) also medial.
- Thus, $C$ (is) also medial [Prop. 10.21].
- And let the (rectangle contained) by $C$ and $D$ be equal to the (square) on $B$.
- And the (square) on $B$ (is) rational.
- Thus, the (rectangle contained) by $C$ and $D$ (is) also rational.
- And since as $A$ is to $B$, so the (rectangle contained) by $A$ and $B$ (is) to the (square) on $B$ [Prop. 10.21 lem.] , but the (square) on $C$ is equal to the (rectangle contained) by $A$ and $B$, and the (rectangle contained) by $C$ and $D$ to the (square) on $B$, thus as $A$ (is) to $B$, so the (square) on $C$ (is) to the (rectangle contained) by $C$ and $D$.
- And as the (square) on $C$ (is) to the (rectangle contained) by $C$ and $D$, so $C$ (is) to $D$ [Prop. 10.21 lem.] .
- And thus as $A$ (is) to $B$, so $C$ (is) to $D$.
- And $A$ is commensurable in square only with $B$.
- Thus, $C$ (is) also commensurable in square only with $D$ [Prop. 10.11].
- And $C$ is medial.
- Thus, $D$ (is) also medial [Prop. 10.23].
- And since as $A$ is to $B$, (so) $C$ (is) to $D$, and the square on $A$ is greater than (the square on) $B$ by the (square) on (some straight line) commensurable (in length) with ($A$), the square on $C$ is thus also greater than (the square on) $D$ by the (square) on (some straight line) commensurable (in length) with ($C$) [Prop. 10.14].
- Thus, two medial (straight lines), $C$ and $D$, commensurable in square only, (and) containing a rational (area), have been found.
- And the square on $C$ is greater than (the square on) $D$ by the (square) on (some straight line) commensurable in length with ($C$).
- So, similarly, (the proposition) can also be demonstrated for (some straight line) incommensurable (in length with $C$), provided that the square on $A$ is greater than (the square on $B$) by the (square) on (some straight line) incommensurable (in length) with ($A$) [Prop. 10.30].
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes
