---
book: 3
number: 15
id: "III.15"
kind: "theorem"
uses: ["[[book-3/proposition-14]]", "[[book-1/proposition-20]]", "[[book-1/proposition-24]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.15"
license: "CC-BY-SA-4.0"
---

# III.15

*Of straight lines in a circle the diameter is greatest*, *and of the rest the nearer to the centre is always greater than the more remote*.

## Proof

Let *ABCD* be a circle, let *AD* be its diameter and *E* the centre; and let *BC* be nearer to the diameter *AD*, and *FG* more remote; I say that *AD* is greatest and *BC* greater than *FG*.

For from the centre *E* let *EH*, *EK* be drawn perpendicular to *BC*, *FG*.

Then, since *BC* is nearer to the centre and *FG* more remote, *EK* is greater than *EH*. [[book-3/definitions#Definition 5|III. Def. 5]]

Let *EL* be made equal to *EH*, through *L* let *LM* be drawn at right angles to *EK* and carried through to *N*, and let *ME*, *EN*, *FE*, *EG* be joined.

Then, since *EH* is equal to *EL*, *BC* is also equal to *MN*. [[book-3/proposition-14|III. 14]]

Again, since *AE* is equal to *EM*, and *ED* to *EN*, *AD* is equal to *ME*, *EN*.

But *ME*, *EN* are greater than *MN*, [[book-1/proposition-20|I. 20]] and *MN* is equal to *BC*; therefore *AD* is greater than *BC*.

And, since the two sides *ME*, *EN* are equal to the two sides *FE*, *EG*, and the angle *MEN* greater than the angle *FEG*, therefore the base *MN* is greater than the base *FG*. [[book-1/proposition-24|I. 24]]

But *MN* was proved equal to *BC*.

Therefore the diameter *AD* is greatest and *BC* greater than *FG*.

Therefore etc. Q. E. D.