## Proposition 29

 In equal circles straight lines that cut off equal circumferences are equal. Let ABC and DEF be equal circles, and in them let equal circumferences BGC and EHF be cut off. Join the straight lines BC and EF. I say that BC equals EF. Take the centers K and L of the circles. Join BK, KC, EL, and LF. III.1 Now, since the circumference BGC equals the circumference EHF, the angle BKC also equals the angle ELF. III.27 And, since the circles ABC and DEF are equal, the radii are also equal, therefore the two sides BK and KC equal the two sides EL and LF, and they contain equal angles, therefore the base BC equals the base EF. I.4 Therefore in equal circles straight lines that cut off equal circumferences are equal. Q.E.D.
This proposition is used in IV.11 and IV.15.

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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