Kitap 6
4
Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE. [I. 28]
4
Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34]
4
And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2]
4
But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16]
4
Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2]
4
But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16]
4
Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]
4
Therefore etc. Q. E. D.
PROPOSITION 5.
5
If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.
5
Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF.
5
For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]
5
Therefore the triangle ABC is equiangular with the triangle GEF.
5
Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4] therefore, as AB is to BC, so is GE to EF.
