PROPOSITION 28. THEOREM.
In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less equal to the less.
Let ABC, DEF be equal circles, and BC,EF equal straight lines in them, which cut off the two greater arcs BAC, EDF, and the two less arcs BGC, EHF the greater arc BAC shall be equal to the greater arc EDF and the less arc BGG equal to the less arc EHF.
Take K, L, the centres of the circles, [III. 1.
and join BK, KC, EL, LF.
Then, because the circles are equal, [Hypothesis. the straight lines from their centres are equal; [III. Def. 1. therefore the two sides BK, KC are equal to the two sides EL, LF, each to each;
and the base BC is equal to the base EF; [Hypothesis.
therefore the angle BKC is equal to the angle ELF. [I. 8.
But in equal circles equal angles stand on equal arcs, when they are at the centres, [III. 26.
therefore the arc BGG is equal to the arc EHF.
But the circumference ABGC is equal to the circumference DEHF; [Hypothesis. therefore the remaining arc BAC is equal to the remaining arc EDF. [Axiom 3.
Wherefore, in equal circles &c. q e.d.
