# Proof: By Euclid

• For let $AG$ be made equal to $E$, and $CH$ equal to $F$.
• In fact, since as $AB$ is to $CD$, so $E$ (is) to $F$, and $E$ (is) equal to $AG$, and $F$ to $CH$, thus as $AB$ is to $CD$, so $AG$ (is) to $CH$.
• And since the whole $AB$ is to the whole $CD$ as the (part) taken away $AG$ (is) to the (part) taken away $CH$, thus the remainder $GB$ will also be to the remainder $HD$ as the whole $AB$ (is) to the whole $CD$ [Prop. 5.19].
• And $AB$ (is) greater than $CD$.
• Thus, $GB$ (is) also greater than $HD$.
• And since $AG$ is equal to $E$, and $CH$ to $F$, thus $AG$ and $F$ is equal to $CH$ and $E$.
• And [since] if [equal (magnitudes) are added to unequal (magnitudes) then the wholes are unequal, thus if] $AG$ and $F$ are added to $GB$, and $CH$ and $E$ to $HD$ - $GB$ and $HD$ being unequal, and $GB$ greater - it is inferred that $AB$ and $F$ (is) greater than $CD$ and $E$.
• Thus, if four magnitudes are proportional then the (sum of the) largest and the smallest of them is greater than the (sum of the) remaining two (magnitudes).
• (Which is) the very thing it was required to show.

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### References

#### Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"