---
book: 2
number: 14
id: "II.14"
kind: "construction"
uses: ["[[book-1/proposition-45]]", "[[book-2/proposition-5]]", "[[book-1/proposition-47]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:2.prop.14"
license: "CC-BY-SA-4.0"
---

# II.14

*To construct a square equal to a given rectilineal figure*.

## Proof

Let *A* be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure *A*.

For let there be constructed the rectangular parallelogram *BD* equal to the rectilineal figure *A*. [[book-1/proposition-45|I. 45]]

Then, if *BE* is equal to *ED*, that which was enjoined will have been done; for a square *BD* has been constructed equal to the rectilineal figure *A*.

But, if not, one of the straight lines *BE*, *ED* is greater.

Let *BE* be greater, and let it be produced to *F*; let *EF* be made equal to *ED*, and let *BF* be bisected at *G*.

With centre *G* and distance one of the straight lines *GB*, *GF* let the semicircle *BHF* be described; let *DE* be produced to *H*, and let *GH* be joined.

Then, since the straight line *BF* has been cut into equal segments at *G*, and into unequal segments at *E*, the rectangle contained by *BE*, *EF* together with the square on *EG* is equal to the square on *GF*. [[book-2/proposition-5|II. 5]]

But *GF* is equal to *GH*; therefore the rectangle *BE*, *EF* together with the square on *GE* is equal to the square on *GH*.

But the squares on *HE*, *EG* are equal to the square on *GH*; [[book-1/proposition-47|I. 47]] therefore the rectangle *BE*, *EF* together with the square on *GE* is equal to the squares on *HE*, *EG*.

Let the square on *GE* be subtracted from each; therefore the rectangle contained by *BE*, *EF* which remains is equal to the square on *EH*.

But the rectangle *BE*, *EF* is *BD*, for *EF* is equal to *ED*; therefore the parallelogram *BD* is equal to the square on *HE*.

And *BD* is equal to the rectilineal figure *A*.

Therefore the rectilineal figure *A* is also equal to the square which can be described on *EH*.

Therefore a square, namely that which can be described on *EH*, has been constructed equal to the given rectilineal figure *A*. Q. E. F.