---
book: 3
number: 35
id: "III.35"
kind: "theorem"
uses: ["[[book-3/proposition-3]]", "[[book-2/proposition-5]]", "[[book-1/proposition-47]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.35"
license: "CC-BY-SA-4.0"
---

# III.35

*If in a circle two straight lines cut one another*, *the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other*.

## Proof

For in the circle *ABCD* let the two straight lines *AC*, *BD* cut one another at the point *E*; I say that the rectangle contained by *AE*, *EC* is equal to the rectangle contained by *DE*, *EB*.

If now *AC*, *BD* are through the centre, so that *E* is the centre of the circle *ABCD*, it is manifest that, *AE*, *EC*, *DE*, *EB* being equal, the rectangle contained by *AE*, *EC* is also equal to the rectangle contained by *DE*, *EB*.

Next let *AC*, *DB* not be through the centre; let the centre of *ABCD* be taken, and let it be *F*; from *F* let *FG*, *FH* be drawn perpendicular to the straight lines *AC*, *DB*, and let *FB*, *FC*, *FE* be joined.

Then, since a straight line *GF* through the centre cuts a straight line *AC* not through the centre at right angles, it also bisects it; [[book-3/proposition-3|III. 3]] therefore *AG* is equal to *GC*.

Since, then, the straight line *AC* has been cut into equal parts at *G* and into unequal parts at *E*, the rectangle contained by *AE*, *EC* together with the square on *EG* is equal to the square on *GC*; [[book-2/proposition-5|II. 5]]

Let the square on *GF* be added; therefore the rectangle *AE*, *EC* together with the squares on *GE*, *GF* is equal to the squares on *CG*, *GF*.

But the square on *FE* is equal to the squares on *EG*, *GF*, and the square on *FC* is equal to the squares on *CG*, *GF*; [[book-1/proposition-47|I. 47]] therefore the rectangle *AE*, *EC* together with the square on *FE* is equal to the square on *FC*.

And *FC* is equal to *FB*; therefore the rectangle *AE*, *EC* together with the square on *EF* is equal to the square on *FB*.

For the same reason, also, the rectangle *DE*, *EB* together with the square on *FE* is equal to the square on *FB*.

But the rectangle *AE*, *EC* together with the square on *FE* was also proved equal to the square on *FB*; therefore the rectangle *AE*, *EC* together with the square on *FE* is equal to the rectangle *DE*, *EB* together with the square on *FE*.

Let the square on *FE* be subtracted from each; therefore the rectangle contained by *AE*, *EC* which remains is equal to the rectangle contained by *DE*, *EB*.

Therefore etc.