[Folio 0062 | p. 56]

PROPOSITION 26.

In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.

[Folio 0063 | p. 57]

Let ABC, DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF; I say that the circumference BKC is equal to the circumference ELF.

For let BC, EF be joined.

Now, since the circles ABC, DEF are equal,

the radii are equal.

Thus the two straight lines BG, GC are equal to the two straight lines EH, HF;

and the angle at G is equal to the angle at H;

[I. 4]therefore the base BC is equal to the base EF.

And, since the angle at A is equal to the angle at D,

[III. Def. 11]the segment BAC is similar to the segment EDF;

and they are upon equal straight lines.

But similar segments of circles on equal straight lines are equal to one another; [III. 24]

therefore the segment BAC is equal to EDF.

But the whole circle ABC is also equal to the whole circle DEF; therefore the circumference BKC which remains is equal to the circumference ELF.

Therefore etc. Q. E. D.

Dependency Graph

Required for

III.27, III.28, III.29, III.30, IV.11, IV.12, IV.15, IV.16, 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.26.