In equal circles equal circumferences are subtended by equal straight lines.
Let ABC, DEF be equal circles, and in them let equal circumferences BGC, EHF
be cut off;
and let the straight lines BC, EF be joined; I say that BC is equal to EF.
For let the centres of the circles be taken, and let them be K, L; let BK, KC,
EL, LF be joined.
Now, since the circumference BGC is equal to the circumference EHF, the angle
BKC is also equal to the angle ELF.
And, since the circles ABC, DEF are equal, the radii are also equal; therefore
the two sides BK, KC are equal to the two sides EL, LF; and they contain equal
angles;
therefore the base BC is equal to the base EF.
Therefore, etc.
Q.E.D.