Proof: By Euclid
(related to Proposition: Prop. 9.15: Sum of Pair of Elements of Geometric Progression with Three Elements in Lowest Terms is Coprime to other Element)
 Let $A$, $B$, $C$ be three numbers in continued proportion (which are) the least of those (numbers) having the same ratio as them.
 I say that two of $A$, $B$, $C$ added together in any way are prime to the remaining (one), (that is) $A$ and $B$ (prime) to $C$, $B$ and $C$ to $A$, and, further, $A$ and $C$ to $B$.
 Let the two least numbers, $DE$ and $EF$, having the same ratio as $A$, $B$, $C$, have been taken [Prop. 8.2].
 So it is clear that $DE$ has made $A$ (by) multiplying itself, and has made $B$ (by) multiplying $EF$, and, further, $EF$ has made $C$ (by) multiplying itself [Prop. 8.2].
 And since $DE$, $EF$ are the least (of those numbers having the same ratio as them), they are prime to one another [Prop. 7.22].
 And if two numbers are prime to one another then the sum (of them) is also prime to each [Prop. 7.28].
 Thus, $DF$ is also prime to each of $DE$, $EF$.
 But, in fact, $DE$ is also prime to $EF$.
 Thus, $DF$, $DE$ are (both) prime to $EF$.
 And if two numbers are (both) prime to some number then the (number) created from (multiplying) them is also prime to the remaining (number) [Prop. 7.24].
 Hence, the (number created) from (multiplying) $FD$, $DE$ is prime to $EF$.
 Hence, the (number created) from (multiplying) $FD$, $DE$ is also prime to the (square) on $EF$ [Prop. 7.25].
 For if two numbers are prime to one another then the (number) created from (squaring) one of them is prime to the remaining (number).] But the (number created) from (multiplying) $FD$, $DE$ is the (square) on $DE$ plus the (number created) from (multiplying) $DE$, $EF$ [Prop. 2.3].
 Thus, the (square) on $DE$ plus the (number created) from (multiplying) $DE$, $EF$ is prime to the (square) on $EF$.
 And the (square) on $DE$ is $A$, and the (number created) from (multiplying) $DE$, $EF$ (is) $B$, and the (square) on $EF$ (is) $C$.
 Thus, $A$, $B$ summed is prime to $C$.
 So, similarly, we can show that $B$, $C$ (summed) is also prime to $A$.
 So I say that $A$, $C$ (summed) is also prime to $B$.
 For since $DF$ is prime to each of $DE$, $EF$ then the (square) on $DF$ is also prime to the (number created) from (multiplying) $DE$, $EF$ [Prop. 7.25].
 But, the (sum of the squares) on $DE$, $EF$ plus twice the (number created) from (multiplying) $DE$, $EF$ is equal to the (square) on $DF$ [Prop. 2.4].
 And thus the (sum of the squares) on $DE$, $EF$ plus twice the (rectangle contained) by $DE$, $EF$ [is] [prime to]bookofproofs$1288 the (rectangle contained) by $DE$, $EF$.
 By separation, the (sum of the squares) on $DE$, $EF$ plus once the (rectangle contained) by $DE$, $EF$ is prime to the (rectangle contained) by $DE$, $EF$.^{1}
 Again, by separation, the (sum of the squares) on $DE$, $EF$ is prime to the (rectangle contained) by $DE$, $EF$.
 And the (square) on $DE$ is $A$, and the (rectangle contained) by $DE$, $EF$ (is) $B$, and the (square) on $EF$ (is) $C$.
 Thus, $A$, $C$ summed is prime to $B$.
 (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