---
book: 6
number: 16
id: "VI.16"
kind: "theorem"
uses: ["[[book-6/proposition-14]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.16"
license: "CC-BY-SA-4.0"
---

# VI.16

*If four straight lines be proportional*, *the rectangle contained by the extremes is equal to the rectangle contained by the means; and*, *if the rectangle contained by the extremes be equal to the rectangle contained by the means*, *the four straight lines will be proportional*.

## Proof

Let the four straight lines *AB*, *CD*, *E*, *F* be proportional, so that, as *AB* is to *CD*, so is *E* to *F*; I say that the rectangle contained by *AB*, *F* is equal to the rectangle contained by *CD*, *E*.

Let *AG*, *CH* be drawn from the points *A*, *C* at right angles to the straight lines *AB*, *CD*, and let *AG* be made equal to *F*, and *CH* equal to *E*.

Let the parallelograms *BG*, *DH* be completed.

Then since, as *AB* is to *CD*, so is *E* to *F*, while *E* is equal to *CH*, and *F* to *AG*, therefore, as *AB* is to *CD*, so is *CH* to *AG*.

Therefore in the parallelograms *BG*, *DH* the sides about the equal angles are reciprocally proportional.

But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [[book-6/proposition-14|VI. 14]] therefore the parallelogram *BG* is equal to the parallelogram *DH*.

And *BG* is the rectangle *AB*, *F*, for *AG* is equal to *F*; and *DH* is the rectangle *CD*, *E*, for *E* is equal to *CH*; therefore the rectangle contained by *AB*, *F* is equal to the rectangle contained by *CD*, *E*.

Next, let the rectangle contained by *AB*, *F* be equal to the rectangle contained by *CD*, *E*; I say that the four straight lines will be proportional, so that, as *AB* is to *CD*, so is *E* to *F*.

For, with the same construction, since the rectangle *AB*, *F* is equal to the rectangle *CD*, *E*, and the rectangle *AB*, *F* is *BG*, for *AG* is equal to *F*, and the rectangle *CD*, *E* is *DH*, for *CH* is equal to *E*, therefore *BG* is equal to *DH*.

And they are equiangular

But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [[book-6/proposition-14|VI. 14]]

Therefore, as *AB* is to *CD*, so is *CH* to *AG*.

But *CH* is equal to *E*, and *AG* to *F*; therefore, as *AB* is to *CD*, so is *E* to *F*.

Therefore etc. Q. E. D.