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Proposition 28

In equal circles equal straight lines cut off equal circumferences, the greater circumference equals the greater and the less equals the less.
Let ABC and DEF be equal circles, and in the circles let AB and DE be equal straight lines cutting off ACB and DFE as greater circumferences and AGB and DHE as lesser. I say that the greater circumference ACB equals the greater circumference DFE, and the less circumference AGB equals DHE.
| Take the centers K and L of the circles, and join AK, KB, DL, and LE. | III.1
Now, since the circles are equal, the radii are also equal, therefore the two sides AK and KB equal the two sides DL and LE, and the base AB equals the base DE, therefore the angle AKB equals the angle DLE. | I.8
But equal angles stand on equal circumferences when they are at the centers, therefore the circumference AGB equals DHE. | III.26
And the whole circle ABC also equals the whole circle DEF, therefore the remaining circumference ACB also equals the remaining circumference DFE.
Therefore in equal circles equal straight lines cut off equal circumferences, the greater circumference equals the greater and the less equals the less.
Q.E.D.

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This proposition is used in III.30 and XIII.18.

Book III Introduction - Proposition III.27 - Proposition III.29.

© 1996
D.E.Joyce
Clark University