Kitap 6
3
But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6]
3
And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE.
3
But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC.
3
Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D.
3
For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE : for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11]
3
Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5]
3
But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.] therefore the angle BAD is also equal to the angle CAD.
3
Therefore the angle BAC has been bisected by the straight line AD.
3
Therefore etc. Q. E. D.
PROPOSITION 4.
4
In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
4
Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
4
For let BC be placed in a straight line with CE.
4
Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet. [I. Post. 5]
4
Let them be produced and meet at F.
4
Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28]
