Kitap 6
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Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4]
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But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB.
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And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH.
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For the same reason the angle CDE is also equal to the angle ABH.
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And the angle at C is also equal to the angle at A, and the angle at E to the angle at H.
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Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1]
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Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE. Q. E. F.
PROPOSITION 19.
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Similar triangles are to one another in the duplicate ratio of the corresponding sides.
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Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF.
