(related to Proposition: 6.16: Rectangles Contained by Proportional Straight Lines)

- Let $AB$, $CD$, $E$, and $F$ be four proportional straight lines, (such that) as $AB$ (is) to $CD$, so $E$ (is) to $F$
- I say that the rectangle contained by $AB$ and $F$ is equal to the rectangle contained by $CD$ and $E$.
- For let $AG$ and $CH$ have been drawn from points $A$ and $C$ at right angles to the straight lines $AB$ and $CD$ (respectively) [Prop. 1.11].
- And let $AG$ be made equal to $F$, and $CH$ to $E$ [Prop. 1.3].
- And let the parallelograms $BG$ and $DH$ have been completed.
- And since as $AB$ is to $CD$, so $E$ (is) to $F$, and $E$ (is) equal $CH$, and $F$ to $AG$, thus as $AB$ is to $CD$, so $CH$ (is) to $AG$.
- Thus, in the parallelograms $BG$ and $DH$ the sides about the equal angles are reciprocally proportional.
- And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.14].
- Thus, parallelogram $BG$ is equal to parallelogram $DH$.
- And $BG$ is the (rectangle contained) by $AB$ and $F$.
- For $AG$ (is) equal to $F$.
- And $DH$ (is) the (rectangle contained) by $CD$ and $E$.
- For $E$ (is) equal to $CH$.
- Thus, the rectangle contained by $AB$ and $F$ is equal to the rectangle contained by $CD$ and $E$.

- And so, let the rectangle contained by $AB$ and $F$ be equal to the rectangle contained by $CD$ and $E$.
- I say that the four straight lines will be proportional, (so that) as $AB$ (is) to $CD$, so $E$ (is) to $F$.
- For, with the same construction, since the (rectangle contained) by $AB$ and $F$ is equal to the (rectangle contained) by $CD$ and $E$.
- And $BG$ is the (rectangle contained) by $AB$ and $F$.
- For $AG$ is equal to $F$.
- And $DH$ (is) the (rectangle contained) by $CD$ and $E$.
- For $CH$ (is) equal to $E$.
- $BG$ is thus equal to $DH$.
- And they are equiangular.
- And in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional [Prop. 6.14].
- Thus, as $AB$ is to $CD$, so $CH$ (is) to $AG$.
- And $CH$ (is) equal to $E$, and $AG$ to $F$.
- Thus, as $AB$ is to $CD$, so $E$ (is) to $F$.
- Thus, if four straight lines are proportional then the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two).
- And if the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two) then the four straight lines will be proportional.

- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"