---
book: 3
number: 7
id: "III.7"
kind: "theorem"
uses: ["[[book-1/proposition-20]]", "[[book-1/proposition-24]]", "[[book-1/proposition-23]]", "[[book-1/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.7"
license: "CC-BY-SA-4.0"
---

# III.7

*If on the diameter of a circle a point be taken which is not the centre of the circle*, *and from the point straight lines fall upon the circle*, *that will be greatest on which the centre is*, *the remainder of the same diameter will be least*, *and of the rest* *the nearer to the straight line through the centre is always greater than the more remote*, *and only two equal straight lines will fall from the point on the circle*, *one on each side of the least straight line*.

## Proof

Let *ABCD* be a circle, and let *AD* be a diameter of it; on *AD* let a point *F* be taken which is not the centre of the circle, let *E* be the centre of the circle, and from *F* let straight lines *FB*, *FC*, *FG* fall upon the circle *ABCD*; I say that *FA* is greatest, *FD* is least, and of the rest *FB* is greater than *FC*, and *FC* than *FG*.

For let *BE*, *CE*, *GE* be joined.

Then, since in any triangle two sides are greater than the remaining one, [[book-1/proposition-20|I. 20]] *EB*, *EF* are greater than *BF*.

But *AE* is equal to *BE*; therefore *AF* is greater than *BF*.

Again, since *BE* is equal to *CE*, and *FE* is common, the two sides *BE*, *EF* are equal to the two sides *CE*, *EF*.

But the angle *BEF* is also greater than the angle *CEF*; therefore the base *BF* is greater than the base *CF*. [[book-1/proposition-24|I. 24]]

For the same reason *CF* is also greater than *FG*.

Again, since *GF*, *FE* are greater than *EG*, and *EG* is equal to *ED*, *GF*, *FE* are greater than *ED*.

Let *EF* be subtracted from each; therefore the remainder *GF* is greater than the remainder *FD*.

Therefore *FA* is greatest, *FD* is least, and *FB* is greater than *FC*, and *FC* than *FG*.

I say also that from the point *F* only two equal straight lines will fall on the circle *ABCD*, one on each side of the least *FD*.

For on the straight line *EF*, and at the point *E* on it, let the angle *FEH* be constructed equal to the angle *GEF* [[book-1/proposition-23|I. 23]], and let *FH* be joined.

Then, since *GE* is equal to *EH*, and *EF* is common, the two sides *GE*, *EF* are equal to the two sides *HE*, *EF*; and the angle *GEF* is equal to the angle *HEF*; therefore the base *FG* is equal to the base *FH*. [[book-1/proposition-4|I. 4]]

I say again that another straight line equal to *FG* will not fall on the circle from the point *F*.

For, if possible, let *FK* so fall.

Then, since *FK* is equal to *FG*, and *FH* to *FG*, *FK* is also equal to *FH*, the nearer to the straight line through the centre being thus equal to the more remote: which is impossible.

Therefore another straight line equal to *GF* will not fall from the point *F* upon the circle; therefore only one straight line will so fall.

Therefore etc. Q. E. D.