---
book: 10
number: 14
id: "X.14"
kind: "theorem"
uses: ["[[book-6/proposition-22]]", "[[book-5/proposition-17]]", "[[book-5/proposition-22]]", "[[book-10/proposition-11]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.14"
license: "CC-BY-SA-4.0"
---

# X.14

*If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by* *the square on a straight line commensurable with the third*.

## Proof

*And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line in-* *commensurable with the third*.

Let *A*, *B*, *C*, *D* be four straight lines in proportion, so that, as *A* is to *B*, so is *C* to *D*; and let the square on *A* be greater than the square on *B* by the square on *E*, and let the square on *C* be greater than the square on *D* by the square on *F*; I say that, if *A* is commensurable with *E*, *C* is also commensurable with *F*, and, if *A* is incommensurable with *E*, *C* is also incommensurable with *F*.

For since, as *A* is to *B*, so is *C* to *D*, therefore also, as the square on *A* is to the square on *B*, so is the square on *C* to the square on *D*. [[book-6/proposition-22|VI. 22]]

But the squares on *E*, *B* are equal to the square on *A*, and the squares on *D*, *F* are equal to the square on *C*.

Therefore, as the squares on *E*, *B* are to the square on *B*, so are the squares on *D*, *F* to the square on *D*; therefore, separando, as the square on *E* is to the square on *B*, so is the square on *F* to the square on *D*; [[book-5/proposition-17|V. 17]] therefore also, as *E* is to *B*, so is *F* to *D*; [[book-6/proposition-22|VI. 22]] therefore, inversely, as *B* is to *E*, so is *D* to *F*.

But, as *A* is to *B*, so also is *C* to *D*; therefore, ex aequali, as *A* is to *E*, so is *C* to *F*. [[book-5/proposition-22|V. 22]]

Therefore, if *A* is commensurable with *E*, *C* is also commensurable with *F*, and, if *A* is incommensurable with *E*, *C* is also incommensurable with *F*. [[book-10/proposition-11|X. 11]]

Therefore etc.