## Proposition 30

 To bisect a given circumference. Let ADB be the given circumference. It is required to bisect the circumference ADB. Join AB, and bisect it at C. Draw CD from the point C at right angles to the straight line AB. Join AD and DB. I.10 I.11 Then, since AC equals CB, and CD is common, the two sides AC and CD equal the two sides BC and CD, and the angle ACD equals the angle BCD, for each is right, therefore the base AD equals the base DB. I.4 But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less, and each of the circumferences AD and DB is less than a semicircle, therefore the circumference AD equals the circumference DB. III.28 Therefore the given circumference has been bisected at the point D. Q.E.F.
The construction in this proposition is used in IV.16.

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 Select from Book III Book III intro III.Def.1 III.Def.2-3 III.Def.4-5 III.Def.6-9 III.Def.10 III.Def.11 III.1 III.2 III.3 III.4 III.5 III.6 III.7 III.8 III.9 III.10 III.11 III.12 III.13 III.14 III.15 III.16 III.17 III.18 III.19 III.20 III.21 III.22 III.23 III.24 III.25 III.26 III.27 III.28 III.29 III.30 III.31 III.32 III.33 III.34 III.35 III.36 III.37 Select book Book I Book II Book III Book IV Book V Book VI Book VII Book VIII Book IX Book X Book XI Book XII Book XIII Select topic Introduction Table of Contents Geometry applet About the text Euclid Web references A quick trip