T H E E L E M E N T S	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. But     EF is equal to BC; therefore     BK is also equal to BC, the less to the greater: which is impossible. Therefore     the angle BGC is not unequal to the angle EHF therefore     it is equal to it. Being (part of) what it was required to prove.
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