Kitap 6
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And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle FGN.
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Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH.
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Therefore, proportionally, as AM is to MB, so is FN to NG, and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH.
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But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as their bases. [VI. 1]
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Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle AMB is to BMC, so is ABE to CBE.
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But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC.
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For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH.
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And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH.
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Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK.
