---
book: 2
number: 5
id: "II.5"
kind: "theorem"
uses: ["[[book-1/proposition-46]]", "[[book-1/proposition-31]]", "[[book-1/proposition-43]]", "[[book-1/proposition-36]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:2.prop.5"
license: "CC-BY-SA-4.0"
---

# II.5

*If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half*.

## Proof

For let a straight line *AB* be cut into equal segments at *C* and into unequal segments at *D*; I say that the rectangle contained by *AD*, *DB* together with the square on *CD* is equal to the square on *CB*.

For let the square *CEFB* be described on *CB*, [[book-1/proposition-46|I. 46]] and let *BE* be joined; through *D* let *DG* be drawn parallel to either *CE* or *BF*, through *H* again let *KM* be drawn parallel to either *AB* or *EF*, and again through *A* let *AK* be drawn parallel to either *CL* or *BM*. [[book-1/proposition-31|I. 31]]

Then, since the complement *CH* is equal to the complement *HF*, [[book-1/proposition-43|I. 43]] let *DM* be added to each; therefore the whole *CM* is equal to the whole *DF*.

But *CM* is equal to *AL*, since *AC* is also equal to *CB*; [[book-1/proposition-36|I. 36]] therefore *AL* is also equal to *DF*. Let *CH* be added to each; therefore the whole *AH* is equal to the gnomon *NOP*.

But *AH* is the rectangle *AD*, *DB*, for *DH* is equal to *DB*, therefore the gnomon *NOP* is also equal to the rectangle *AD*, *DB*.

Let *LG*, which is equal to the square on *CD*, be added to each; therefore the gnomon *NOP* and *LG* are equal to the rectangle contained by *AD*, *DB* and the square on *CD*.

But the gnomon *NOP* and *LG* are the whole square *CEFB*, which is described on *CB*; therefore the rectangle contained by *AD*, *DB* together with the square on *CD* is equal to the square on *CB*.

Therefore etc. Q. E. D.