Proof: By Euclid
(related to Proposition: Prop. 10.028: Construction of Components of Second Bimedial)
- Let the [three] [rational]bookofproofs$2083 (straight lines) $A$, $B$, and $C$, (which are) commensurable in square only, be laid down.
- And let, $D$, in mean proportion3 (straight line) to $A$ and $B$, have been taken [Prop. 6.13].
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And let it be contrived that as $B$ (is) to $C$, (so) $D$ (is) to $E$ [Prop. 6.12].
-
Since the rational (straight lines) $A$ and $B$ are commensurable in square only, the (rectangle contained) by $A$ and $B$ - that is to say, the (square) on $D$ [Prop. 6.17] - is medial [Prop. 10.21].
- Thus, $D$ (is) medial [Prop. 10.21].
- And since $B$ and $C$ are commensurable in square only, and as $B$ is to $C$, (so) $D$ (is) to $E$, $D$ and $E$ are thus commensurable in square only [Prop. 10.11].
- And $D$ (is) medial.
- $E$ (is) thus also medial [Prop. 10.23].
- Thus, $D$ and $E$ are medial (straight lines which are) commensurable in square only.
- So, I say that they also enclose a medial (area) .
- For since as $B$ is to $C$, (so) $D$ (is) to $E$, thus, alternately, as $B$ (is) to $D$, (so) $C$ (is) to $E$ [Prop. 5.16].
- And as $B$ (is) to $D$, (so) $D$ (is) to $A$.
- And thus as $D$ (is) to $A$, (so) $C$ (is) to $E$.
- Thus, the (rectangle contained) by $A$ and $C$ is equal to the (rectangle contained) by $D$ and $E$ [Prop. 6.16].
- And the (rectangle contained) by $A$ and $C$ is medial [Prop. 10.21].
- Thus, the (rectangle contained) by $D$ and $E$ (is) also medial.
- Thus, (two) medial (straight lines, $D$ and $E$), containing a medial (area) , (which are) commensurable in square only, have been found.
- (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
