[Folio 0064 | p. 58]

PROPOSITION 27.

In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.

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.

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]

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.

But EF is equal to BC;

therefore BK is also equal to BC, the less to the greater: which is impossible.

Therefore the angle BGC is not unequal to the angle EHF;

therefore it is equal to it.

[Folio 0065 | p. 59]

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.

Therefore etc. Q. E. D.

Dependency Graph

Required for

III.29, IV.11, IV.12, IV.15, VI.33, XII.1, XII.2, XII.5, XII.6, XII.7, XII.8, XII.10, XII.11, XII.12, XII.13, XII.14, XII.15, XII.17, XII.18, XIII.8, XIII.9, XIII.10, XIII.11, XIII.12, XIII.13, XIII.16, XIII.18

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See Commentary on Proposition III.27.