Kitap 11
23
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.
23
Then, since AB is equal to BC, OP is also equal to OQ, so that the remainder LP is equal to QM.
23
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]
23
But LO is greater than OP; therefore LM is also greater than PQ.
23
But LM was made equal to AC; therefore AC is also greater than PQ.
23
Since, then, the two sides AB, BC are equal to the two sides PO, OQ, and the base AC is greater than the base PQ, therefore the angle ABC is greater than the angle POQ. [I. 25]
23
Similarly we can prove that the angle DEF is also greater than the angle MON, and the angle GHK greater than the angle NOL.
23
Therefore the three angles ABC, DEF, GHK are greater than the three angles LOM, MON, NOL.
23
But, by hypothesis, the angles ABC, DEF, GHK are less than four right angles; therefore the angles LOM, MON, NOL are much less than four right angles.
23
But they are also equal to four right angles: which is absurd.
23
Therefore AB is not less than LO.
23
And it was proved that neither is it equal; therefore AB is greater than LO.
23
Let then OR be set up from the point O at right angles to the plane of the circle LMN, [XI. 12] and let the square on OR be equal to that area by which the square on AB is greater than the square on LO; [Lemma] let RL, RM, RN be joined.
23
Then, since RO is at right angles to the plane of the circle LMN, therefore RO is also at right angles to each of the straight lines LO, MO, NO.
23
And, since LO is equal to OM, while OR is common and at right angles, therefore the base RL is equal to the base RM. [I. 4]
23
For the same reason RN is also equal to each of the straight lines RL, RM; therefore the three straight lines RL, RM, RN are equal to one another.
