Kitap 11
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Similarly we can prove that AC, DF are greater than GK, and further DF, GK are greater than AC.
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Therefore it is possible to construct a triangle out of straight lines equal to AC, DF, GK. Q. E. D.
PROPOSITION 23.
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To construct a solid angle out of three plane angles two of which, taken together in any manner, are greater than the remaining one: thus the three angles must be less than four right angles.
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Let the angles ABC, DEF, GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles ABC, DEF, GHK.
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Let AB, BC, DE, EF, GH, HK be cut off equal to one another, and let AC, DF, GK be joined; it is therefore possible to construct a triangle out of straight lines equal to AC, DF, GK. [XI. 22]
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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.
