---
book: 6
number: 17
id: "VI.17"
kind: "theorem"
uses: ["[[book-6/proposition-16]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.17"
license: "CC-BY-SA-4.0"
---

# VI.17

*If three straight lines be proportional*, *the rectangle contained by the extremes is equal to the square on the mean*; *and*, *if the rectangle contained by the extremes be equal to the square on the mean*, *the three straight lines will be proportional*.

## Proof

Let the three straight lines *A*, *B*, *C* be proportional, so that, as *A* is to *B*, so is *B* to *C*; I say that the rectangle contained by *A*, *C* is equal to the square on *B*.

Let *D* be made equal to *B*.

Then, since, as *A* is to *B*, so is *B* to *C*, and *B* is equal to *D*, therefore, as *A* is to *B*, so is *D* to *C*.

But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [[book-6/proposition-16|VI. 16]]

Therefore the rectangle *A*, *C* is equal to the rectangle *B*, *D*.

But the rectangle *B*, *D* is the square on *B*, for *B* is equal to *D*; therefore the rectangle contained by *A*, *C* is equal to the square on *B*.

Next, let the rectangle *A*, *C* be equal to the square on *B*; I say that, as *A* is to *B*, so is *B* to *C*.

For, with the same construction, since the rectangle *A*, *C* is equal to the square on *B*, while the square on *B* is the rectangle *B*, *D*, for *B* is equal to *D*, therefore the rectangle *A*, *C* is equal to the rectangle *B*, *D*.

But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [[book-6/proposition-16|VI. 16]]

Therefore, as *A* is to *B*, so is *D* to *C*.

But *B* is equal to *D*; therefore, as *A* is to *B*, so is *B* to *C*.

Therefore etc. Q. E. D.