---
book: 10
number: 46
id: "X.46"
kind: "theorem"
uses: ["[[book-10/proposition-40]]", "[[book-10/proposition-26]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.46"
license: "CC-BY-SA-4.0"
---

# X.46

*The side of a rational plus a medial area is divided at one point only*.

## Proof

Let *AB* be the side of a rational plus a medial area divided at *C*, so that *AC*, *CB* are incommensurable in square and make the sum of the squares on *AC*, *CB* medial, but twice the rectangle *AC*, *CB* rational; [[book-10/proposition-40|X. 40]] I say that *AB* is not so divided at another point.

For, if possible, let it be divided at *D* also, so that *AD*, *DB* are also incommensurable in square and make the sum of the squares on *AD*, *DB* medial, but twice the rectangle *AD*, *DB* rational.

Since then that by which twice the rectangle *AC*, *CB* differs from twice the rectangle *AD*, *DB* is also that by which the squares on *AD*, *DB* differ from the squares on *AC*, *CB*, while twice the rectangle *AC*, *CB* exceeds twice the rectangle *AD*, *DB* by a rational area, therefore the squares on *AD*, *DB* also exceed the squares on *AC*, *CB* by a rational area, though they are medial: which is impossible. [[book-10/proposition-26|X. 26]]

Therefore the side of a rational plus a medial area is not divided at different points; therefore it is divided at one point only. Q. E. D.