---
book: 10
number: 17
id: "X.17"
kind: "theorem"
uses: ["[[book-2/proposition-5]]", "[[book-10/proposition-15]]", "[[book-10/proposition-6]]", "[[book-10/proposition-12]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.17"
license: "CC-BY-SA-4.0"
---

# X.17

*If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater.*

## Proof

*And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with* *the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length*.

Let *A*, *BC* be two unequal straight lines, of which *BC* is the greater, and let there be applied to *BC* a parallelogram equal to the fourth part of the square on the less, *A*, that is, equal to the square on the half of *A*, and deficient by a square figure. Let this be the rectangle *BD*, *DC*, [cf. Lemma] and let *BD* be commensurable in length with *DC*; I say that the square on *BC* is greater than the square on *A* by the square on a straight line commensurable with *BC*.

For let *BC* be bisected at the point *E*, and let *EF* be made equal to *DE*.

Therefore the remainder *DC* is equal to *BF*.

And, since the straight line *BC* has been cut into equal parts at *E*, and into unequal parts at *D*, therefore the rectangle contained by *BD*, *DC*, together with the square on *ED*, is equal to the square on *EC*; [[book-2/proposition-5|II. 5]]

And the same is true of their quadruples; therefore four times the rectangle *BD*, *DC*, together with four times the square on *DE*, is equal to four times the square on *EC*.

But the square on *A* is equal to four times the rectangle *BD*, *DC*; and the square on *DF* is equal to four times the square on *DE*, for *DF* is double of *DE*.

And the square on *BC* is equal to four times the square on *EC*, for again *BC* is double of *CE*.

Therefore the squares on *A*, *DF* are equal to the square on *BC*, so that the square on *BC* is greater than the square on *A* by the square on *DF*.

It is to be proved that *BC* is also commensurable with *DF*.

Since *BD* is commensurable in length with *DC*, therefore *BC* is also commensurable in length with *CD*. [[book-10/proposition-15|X. 15]]

But *CD* is commensurable in length with *CD*, *BF*, for *CD* is equal to *BF*. [[book-10/proposition-6|X. 6]]

Therefore *BC* is also commensurable in length with *BF*, *CD*, [[book-10/proposition-12|X. 12]] so that *BC* is also commensurable in length with the remainder *FD*; [[book-10/proposition-15|X. 15]] therefore the square on *BC* is greater than the square on *A* by the square on a straight line commensurable with *BC*.

Next, let the square on *BC* be greater than the square on *A* by the square on a straight line commensurable with *BC*, let a parallelogram be applied to *BC* equal to the fourth part of the square on *A* and deficient by a square figure, and let it be the rectangle *BD*, *DC*.

It is to be proved that *BD* is commensurable in length with *DC*.

With the same construction, we can prove similarly that the square on *BC* is greater than the square on *A* by the square on *FD*.

But the square on *BC* is greater than the square on *A* by the square on a straight line commensurable with *BC*.

Therefore *BC* is commensurable in length with *FD*, so that *BC* is also commensurable in length with the remainder, the sum of *BF*, *DC*. [[book-10/proposition-15|X. 15]]

But the sum of *BF*, *DC* is commensurable with *DC*, [[book-10/proposition-6|X. 6]] so that *BC* is also commensurable in length with *CD*; [[book-10/proposition-12|X. 12]] and therefore, separando, *BD* is commensurable in length with *DC*. [[book-10/proposition-15|X. 15]]

Therefore etc.