Kitap 6
6
But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11]
6
Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]
6
Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF.
6
But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE.
6
And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF.
6
Therefore etc. Q. E. D.
PROPOSITION 7.
7
If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.
7
Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F.
7
For, if the angle ABC is unequal to the angle DEF, one of them is greater.
7
Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23]
