---
book: 10
number: 47
id: "X.47"
kind: "theorem"
uses: ["[[book-2/proposition-4]]", "[[book-10/proposition-22]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-36]]", "[[book-10/proposition-42]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.47"
license: "CC-BY-SA-4.0"
---

# X.47

*The side of the sum of two medial areas is divided at one point only*.

## Proof

Let *AB* be divided at *C*, so that *AC*, *CB* are incommensurable in square and make the sum of the squares on *AC*, *CB* medial, and the rectangle *AC*, *CB* medial and also incommensurable with the sum of the squares on them; I say that *AB* is not divided at another point so as to fulfil the given conditions.

For, if possible, let it be divided at *D*, so that again *AC* is of course not the same as *BD*, but *AC* is supposed greater; let a rational straight line *EF* be set out, and let there be applied to *EF* the rectangle *EG* equal to the squares on *AC*, *CB*, and the rectangle *HK* equal to twice the rectangle *AC*, *CB*; therefore the whole *EK* is equal to the square on *AB*. [[book-2/proposition-4|II. 4]]

Again, let *EL*, equal to the squares on *AD*, *DB*, be applied to *EF*; therefore the remainder, twice the rectangle *AD*, *DB*, is equal to the remainder *MK*.

And since, by hypothesis, the sum of the squares on *AC*, *CB* is medial, therefore *EG* is also medial.

And it is applied to the rational straight line *EF*; therefore *HE* is rational and incommensurable in length with *EF*. [[book-10/proposition-22|X. 22]]

For the same reason *HN* is also rational and incommensurable in length with *EF*.

And, since the sum of the squares on *AC*, *CB* is incommensurable with twice the rectangle *AC*, *CB*, therefore *EG* is also incommensurable with *GN*, so that *EH* is also incommensurable with *HN*. [[book-6/proposition-1|VI. 1]] , [[book-10/proposition-11|X. 11]]

And they are rational; therefore *EH*, *HN* are rational straight lines commensurable in square only; therefore *EN* is a binomial straight line divided at *H*. [[book-10/proposition-36|X. 36]]

Similarly we can prove that it is also divided at *M.*

And *EH* is not the same with *MN*; therefore a binomial has been divided at different points: which is absurd. [[book-10/proposition-42|X. 42]]

Therefore a side of the sum of two medial areas is not divided at different points; therefore it is divided at one point only.