---
book: 2
number: 11
id: "II.11"
kind: "construction"
uses: ["[[book-1/proposition-46]]", "[[book-2/proposition-6]]", "[[book-1/proposition-47]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:2.prop.11"
license: "CC-BY-SA-4.0"
---

# II.11

*To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment*.

## Proof

Let *AB* be the given straight line; thus it is required to cut *AB* so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

For let the square *ABDC* be described on *AB*; [[book-1/proposition-46|I. 46]] let *AC* be bisected at the point *E*, and let *BE* be joined; let *CA* be drawn through to *F*, and let *EF* be made equal to *BE*; let the square *FH* be described on *AF*, and let *GH* be drawn through to *K*.

I say that *AB* has been cut at *H* so as to make the rectangle contained by *AB*, *BH* equal to the square on *AH*.

For, since the straight line *AC* has been bisected at *E*, and *FA* is added to it, the rectangle contained by *CF*, *FA* together with the square on *AE* is equal to the square on *EF*. [[book-2/proposition-6|II. 6]]

But *EF* is equal to *EB*; therefore the rectangle *CF*, *FA* together with the square on *AE* is equal to the square on *EB*.

But the squares on *BA*, *AE* are equal to the square on *EB*, for the angle at *A* is right; [[book-1/proposition-47|I. 47]] therefore the rectangle *CF*, *FA* together with the square on *AE* is equal to the squares on *BA*, *AE*.

Let the square on *AE* be subtracted from each; therefore the rectangle *CF*, *FA* which remains is equal to the square on *AB*.

Now the rectangle *CF*, *FA* is *FK*, for *AF* is equal to *FG*; and the square on *AB* is *AD*; therefore *FK* is equal to *AD*.

Let *AK* be subtracted from each; therefore *FH* which remains is equal to *HD*.

And *HD* is the rectangle *AB*, *BH*, for *AB* is equal to *BD*; and *FH* is the square on *AH*; therefore the rectangle contained by *AB*, *BH* is equal to the square on *HA*. therefore the given straight line *AB* has been cut at *H* so as to make the rectangle contained by *AB*, *BH* equal to the square on *HA*. Q. E. F.