Book 3
27
For in equal circles ABC, DEF, on equal circumferences BC, EF, let the angles BGC, EHF stand at the centres G, H, and the angles BAC, EDF at the circumferences; I say that the angle BGC is equal to the angle EHF, and the angle BAC is equal to the angle EDF.
27
For, if the angle BGC is unequal to the angle EHF, one of them is greater. Let the angle BGC be greater : and on the straight line BG, and at the point G on it, let the angle BGK be constructed equal to the angle EHF. [I. 23]
27
Now equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference BK is equal to the circumference EF.
27
But EF is equal to BC; therefore BK is also equal to BC, the less to the greater : which is impossible.
27
Therefore the angle BGC is not unequal to the angle EHF; therefore it is equal to it.
27
And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; [III. 20] therefore the angle at A is also equal to the angle at D.
27
Therefore etc. Q. E. D.
PROPOSITION 28.
28
In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less.
28
Let ABC, DEF be equal circles, and in the circles let AB, DE be equal straight lines cutting off ACB, DFE as greater circumferences and AGB, DHE as lesser; I say that the greater circumference ACB is equal to the greater circumference DFE, and the less circumference AGB to DHE.
28
For let the centres K, L of the circles be taken, and let AK, KB, DL, LE be joined.
28
Now, since the circles are equal, the radii are also equal; therefore the two sides AK, KB are equal to the two sides DL, LE; and the base AB is equal to the base DE; therefore the angle AKB is equal to the angle DLE. [I. 8]
28
But equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference AGB is equal to DHE.
