Book 12
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Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, SE; therefore the remainder, the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex, is greater than the solid O.
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Let there be also inscribed in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS, and let there be set up on the polygon ATBUCVDW a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon ATBUCVDW is the base and the point L the vertex let LBT be one, and of the triangles containing the pyramid of which the polygon EPFQGRHS is the base and the point N the vertex let NFP be one; and let KT, MP be joined.
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Now, since the cone ABCDL is similar to the cone EFGHN, therefore, as BD is to FH, so is the axis KL to the axis MN. [XI. Def. 24]
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But, as BD is to FH, so is BK to FM; therefore also, as BK is to FM, so is KL to MN.
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And, alternately, as BK is to KL, so is FM to MN. [V. 16]
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And the sides are proportional about equal angles, namely the angles BKL, FMN; therefore the triangle BKL is similar to the triangle FMN. [VI. 6]
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Again, since, as BK is to KT, so is FM to MP, and they are about equal angles, namely the angles BKT, FMP, inasmuch as, whatever part the angle BKT is of the four right angles at the centre K, the same part also is the angle FMP of the four right angles at the centre M; since then the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP. [VI. 6]
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Again, since it was proved that, as BK is to KL, so is FM to MN, while BK is equal to KT, and FM to PM, therefore, as TK is to KL, so is PM to MN; and the sides are proportional about equal angles, namely the angles TKL, PMN, for they are right; therefore the triangle LKT is similar to the triangle NMP. [VI. 6]
