---
book: 2
number: 4
id: "II.4"
kind: "theorem"
uses: ["[[book-1/proposition-46]]", "[[book-1/proposition-31]]", "[[book-1/proposition-29]]", "[[book-1/proposition-5]]", "[[book-1/proposition-6]]", "[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:2.prop.4"
license: "CC-BY-SA-4.0"
---

# II.4

*If a straight line be cut at random*, *the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments*.

## Proof

For let the straight line *AB* be cut at random at *C*; I say that the square on *AB* is equal to the squares on *AC*, *CB* and twice the rectangle contained by *AC*, *CB*.

For let the square *ADEB* be described on *AB*, [[book-1/proposition-46|I. 46]] let *BD* be joined; through *C* let *CF* be drawn parallel to either *AD* or *EB*, and through *G* let *HK* be drawn parallel to either *AB* or *DE*. [[book-1/proposition-31|I. 31]]

Then, since *CF* is parallel to *AD*, and *BD* has fallen on them, the exterior angle *CGB* is equal to the interior and opposite angle *ADB*. [[book-1/proposition-29|I. 29]]

But the angle *ADB* is equal to the angle *ABD*, since the side *BA* is also equal to *AD*; [[book-1/proposition-5|I. 5]] therefore the angle *CGB* is also equal to the angle *GBC*, so that the side *BC* is also equal to the side *CG*. [[book-1/proposition-6|I. 6]]

But *CB* is equal to *GK*, and *CG* to *KB*; [[book-1/proposition-34|I. 34]] therefore *GK* is also equal to *KB*; therefore *CGKB* is equilateral.

I say next that it is also right-angled.

For, since *CG* is parallel to *BK*, the angles *KBC*, *GCB* are equal to two right angles. [[book-1/proposition-29|I. 29]]

But the angle *KBC* is right; therefore the angle *BCG* is also right, so that the opposite angles *CGK*, *GKB* are also right. [[book-1/proposition-34|I. 34]]

Therefore *CGKB* is right-angled; and it was also proved equilateral; therefore it is a square; and it is described on *CB*.

For the same reason *HF* is also a square; and it is described on *HG*, that is *AC*. [[book-1/proposition-34|I. 34]]

Therefore the squares *HF*, *KC* are the squares on *AC*, *CB*.

Now, since *AG* is equal to *GE*, and *AG* is the rectangle *AC*, *CB*, for *GC* is equal to *CB*, therefore *GE* is also equal to the rectangle *AC*, *CB*.

Therefore *AG*, *GE* are equal to twice the rectangle *AC*, *CB*.

But the squares *HF*, *CK* are also the squares on *AC*, *CB*; therefore the four areas *HF*, *CK*, *AG*, *GE* are equal to the squares on *AC*, *CB* and twice the rectangle contained by *AC*, *CB*.

But *HF*, *CK*, *AG*, *GE* are the whole *ADEB*, which is the square on *AB*.

Therefore the square on *AB* is equal to the squares on *AC*, *CB* and twice the rectangle contained by *AC*, *CB*.

Therefore etc. Q. E. D.