---
book: 10
number: 32
id: "X.32"
kind: "construction"
uses: ["[[book-10/proposition-29]]", "[[book-10/proposition-21]]", "[[book-10/proposition-11]]", "[[book-10/proposition-23]]", "[[book-10/proposition-14]]", "[[book-10/proposition-30]]", "[[book-6/proposition-8]]", "[[book-6/proposition-4]]", "[[book-6/proposition-17]]", "[[book-6/proposition-16]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.32"
license: "CC-BY-SA-4.0"
---

# X.32

*To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.*

## Proof

Let there be set out three rational straight lines *A*, *B*, *C* commensurable in square only, and such that the square on *A* is greater than the square on *C* by the square on a straight line commensurable with *A*, [[book-10/proposition-29|X. 29]] and let the square on *D* be equal to the rectangle *A*, *B*.

Therefore the square on *D* is medial; therefore *D* is also medial. [[book-10/proposition-21|X. 21]]

Let the rectangle *D*, *E* be equal to the rectangle *B*, *C*.

Then since, as the rectangle *A*, *B* is to the rectangle *B*, *C*, so is *A* to *C*; while the square on *D* is equal to the rectangle *A*, *B*, and the rectangle *D*, *E* is equal to the rectangle *B*, *C*, therefore, as *A* is to *C*, so is the square on *D* to the rectangle *D*, *E*.

But, as the square on *D* is to the rectangle *D*, *E*, so is *D* to *E*; therefore also, as *A* is to *C*, so is *D* to *E*.

But *A* is commensurable with *C* in square only; therefore *D* is also commensurable with *E* in square only. [[book-10/proposition-11|X. 11]]

But *D* is medial; therefore *E* is also medial. [[book-10/proposition-23|X. 23]], addition

And, since, as *A* is to *C*, so is *D* to *E*, while the square on *A* is greater than the square on *C* by the square on a straight line commensurable with *A*, therefore also the square on *D* will be greater than the square on *E* by the square on a straight line commensurable with *D*.[[book-10/proposition-14|X. 14]]

I say next that the rectangle *D*, *E* is also medial.

For, since the rectangle *B*, *C* is equal to the rectangle *D*, *E*, while the rectangle *B*, *C* is medial, [[book-10/proposition-21|X. 21]] therefore the rectangle *D*, *E* is also medial.

Therefore two medial straight lines *D*, *E*, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

Similarly again it can be proved that the square on *D* is greater than the square on *E* by the square on a straight line incommensurable with *D*, when the square on *A* is greater than the square on *C* by the square on a straight line incommensurable with *A*. [[book-10/proposition-30|X. 30]]

Lemma. Let *ABC* be a right-angled triangle having the angle *A* right, and let the perpendicular *AD* be drawn; I say that the rectangle *CB*, *BD* is equal to the square on *BA*, the rectangle *BC*, *CD* equal to the square on *CA*, the rectangle *BD*, *DC* equal to the square on *AD*, and, further, the rectangle *BC*, *AD* equal to the rectangle *BA*, *AC*.

And first that the rectangle *CB*, *BD* is equal to the square on *BA*.

For, since in a right-angled triangle *AD* has been drawn from the right angle perpendicular to the base, therefore the triangles *ABD*, *ADC* are similar both to the whole *ABC* and to one another. [[book-6/proposition-8|VI. 8]]

And since the triangle *ABC* is similar to the triangle *ABD*, therefore, as *CB* is to *BA*, so is *BA* to *BD*; [[book-6/proposition-4|VI. 4]] therefore the rectangle *CB*, *BD* is equal to the square on *AB*. [[book-6/proposition-17|VI. 17]]

For the same reason the rectangle *BC*, *CD* is also equal to the square on *AC*.

And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [[book-6/proposition-8|VI. 8, Por.]] therefore, as *BD* is to *DA*, so is *AD* to *DC*; therefore the rectangle *BD*, *DC* is equal to the square on *AD*. [[book-6/proposition-17|VI. 17]]

I say that the rectangle *BC*, *AD* is also equal to the rectangle *BA*, *AC*.

For since, as we said, *ABC* is similar to *ABD*, therefore, as *BC* is to *CA*, so is *BA* to *AD*. [[book-6/proposition-4|VI. 4]]

Therefore the rectangle *BC*, *AD* is equal to the rectangle *BA*, *AC*. [[book-6/proposition-16|VI. 16]] Q. E. D.