---
book: 3
number: 30
id: "III.30"
kind: "construction"
uses: ["[[book-1/proposition-4]]", "[[book-3/proposition-28]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.30"
license: "CC-BY-SA-4.0"
---

# III.30

*To bisect a given circumference*.

## Proof

Let *ADB* be the given circumference; thus it is required to bisect the circumference *ADB*.

Let *AB* be joined and bisected at *C*; from the point *C* let *CD* be drawn at right angles to the straight line *AB*, and let *AD*, *DB* be joined.

Then, since *AC* is equal to *CB*, and *CD* is common, the two sides *AC*, *CD* are equal to the two sides *BC*, *CD*; and the angle *ACD* is equal to the angle *BCD*, for each is right; therefore the base *AD* is equal to the base *DB*. [[book-1/proposition-4|I. 4]]

But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [[book-3/proposition-28|III. 28]] and each of the circumferences *AD*, *DB* is less than a semicircle; therefore the circumference *AD* is equal to the circumference *DB*.

Therefore the given circumference has been bisected at the point *D*. Q. E. F.