|Book III Proposition 27|
Now equal angles stand on equal circumferences, when thay are at the centers; [III.26] therefore
the circumference BK is equal to the circumference EF.
EF is equal to BC;
BK is also equal to BC,
the angle BGC is not unequal to the angle EHF
it is equal to it.
Being (part of) what it was required to prove.
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