Kitap 11
23
Let LMN be so constructed that AC is equal to LM, DF to MN, and further GK to NL, let the circle LMN be described about the triangle LMN, let its centre be taken, and let it be O; let LO, MO, NO be joined; I say that AB is greater than LO.
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For, if not, AB is either equal to LO, or less.
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First, let it be equal.
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Then, since AB is equal to LO, while AB is equal to BC, and OL to OM, the two sides AB, BC are equal to the two sides LO, OM respectively; and, by hypothesis, the base AC is equal to the base LM; therefore the angle ABC is equal to the angle LOM. [I. 8]
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For the same reason the angle DEF is also equal to the angle MON, and further the angle GHK to the angle NOL; therefore the three angles ABC, DEF, GHK are equal to the three angles LOM, MON, NOL.
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But the three angles LOM, MON, NOL are equal to four right angles; therefore the angles ABC, DEF, GHK are equal to four right angles.
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But they are also, by hypothesis, less than four right angles: which is absurd.
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Therefore AB is not equal to LO.
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I say next that neither is AB less than LO.
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For, if possible, let it be so, and let OP be made equal to AB, and OQ equal to BC, and let PQ be joined.
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Then, since AB is equal to BC, OP is also equal to OQ, so that the remainder LP is equal to QM.
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Therefore LM is parallel to PQ, [VI. 2] and LMO is equiangular with PQO; [I. 29] therefore, as OL is to LM, so is OP to PQ; [VI. 4] and alternately, as LO is to OP, so is LM to PQ. [V. 16]
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But LO is greater than OP; therefore LM is also greater than PQ.
