Proof: By Euclid
(related to Proposition: 5.11: Equality of Ratios is Transitive)
- For let the equal multiples $G$, $H$, $K$ have been taken of $A$, $C$, $E$ (respectively), and the other random equal multiples $L$, $M$, $N$ of $B$, $D$, $F$ (respectively).
- And since as $A$ is to $B$, so $C$ (is) to $D$, and the equal multiples $G$ and $H$ have been taken of $A$ and $C$ (respectively), and the other random equal multiples $L$ and $M$ of $B$ and $D$ (respectively), thus if $G$ exceeds $L$ then $H$ also exceeds $M$, and if ($G$ is) equal (to $L$ then $H$ is also) equal (to $M$), and if ($G$ is) less (than $L$ then $H$ is also) less (than $M$) [Def. 5.5] .
- Again, since as $C$ is to $D$, so $E$ (is) to $F$, and the equal multiples $H$ and $K$ have been taken of $C$ and $E$ (respectively), and the other random equal multiples $M$ and $N$ of $D$ and $F$ (respectively), thus if $H$ exceeds $M$ then $K$ also exceeds $N$, and if ($H$ is) equal (to $M$ then $K$ is also) equal (to $N$), and if ($H$ is) less (than $M$ then $K$ is also) less (than $N$) [Def. 5.5] .
- But (we saw that) if $H$ was exceeding $M$ then $G$ was also exceeding $L$, and if ($H$ was) equal (to $M$ then $G$ was also) equal (to $L$), and if ($H$ was) less (than $M$ then $G$ was also) less (than $L$).
- And, hence, if $G$ exceeds $L$ then $K$ also exceeds $N$, and if ($G$ is) equal (to $L$ then $K$ is also) equal (to $N$), and if ($G$ is) less (than $L$ then $K$ is also) less (than $N$).
- And $G$ and $K$ are equal multiples of $A$ and $E$ (respectively), and $L$ and $N$ other random equal multiples of $B$ and $F$ (respectively).
- Thus, as $A$ is to $B$, so $E$ (is) to $F$ [Def. 5.5] .
- Thus, (ratios which are) the same with the same ratio are also the same with one another.
- (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"
