---
book: 8
number: 14
id: "VIII.14"
kind: "theorem"
uses: ["[[book-8/proposition-11]]", "[[book-8/proposition-7]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:8.prop.14"
license: "CC-BY-SA-4.0"
---

# VIII.14

*If a square measure a square, the side will also measure the side; and, if the side measure the side, the square will also measure the square.*

## Proof

Let *A*, *B* be square numbers, let *C*, *D* be their sides, and let *A* measure *B*; I say that *C* also measures *D*.

For let *C* by multiplying *D* make *E*; therefore *A*, *E*, *B* are continuously proportional in the ratio of *C* to *D*. [[book-8/proposition-11|VIII. 11]]

And, since *A*, *E*, *B* are continuously proportional, and *A* measures *B*, therefore *A* also measures *E*. [[book-8/proposition-7|VIII. 7]]

And, as *A* is to *E*, so is *C* to *D*; therefore also *C* measures *D*. [[book-7/definitions#Definition 20|VII. Def. 20]]

Again, let *C* measure *D*; I say that *A* also measures *B*.

For, with the same construction, we can in a similar manner prove that *A*, *E*, *B* are continuously proportional in the ratio of *C* to *D*.

And since, as *C* is to *D*, so is *A* to *E*, and *C* measures *D*, therefore *A* also measures *E*. [[book-7/definitions#Definition 20|VII. Def. 20]]

And *A*, *E*, *B* are continuously proportional; therefore *A* also measures *B*.

Therefore etc. Q. E. D.