PROPOSITION 26. THEOREM.
In equal circles, equal angles stand on equal arcs, whether they be at the centres or circumferences.
Let ABC, DEF be equal circles; and let BGC, EHF be equal angles in them at their centres, and BAC, EDF equal angles at their circumferences: the arc BKC shall be equal to the arc ELF.
Join BC, EF.
Then, because the circles ABC, DEF are equal, [Hyp.
the straight lines from their centres arc equal; [III. Def. 1.
therefore the two sides BG, GC are equal to the two sides EH, HF, each to each;
and the angle at G is equal to the angle at H; [Hypothesis.
therefore the base BC is equal to the base EF. [I. 4.
And because the angle at A is equal to the angle at D,[Hyp.
the segment BAC dissimilar to the segment EDF; [III. Def. ll.
and they are on equal straight lines BC, EF.
But similar segments of circles on equal straight lines are equal to one another; [III. 24.
therefore the segment BAC be equal to the segment EDF.
But the whole circle ABC is equal to the whole circle DEF; [Hypothesis.
therefore the remaining segment BKC is equal to the remaining segment ELF; [Axiom 3.
therefore the arc BKC is equal to the arc ELF.
Wherefore, in equal circies &c. q.e.d.
