Kitap 6
2
For the same reason also, as the triangle CDE is to ADE, so is CE to EA.
2
Therefore also, as BD is to DA, so is CE to EA. [V. 11]
2
Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined.
2
I say that DE is parallel to BC.
2
For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11]
2
Therefore each of the triangles BDE, CDE has the same ratio to ADE.
2
Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE.
2
But equal triangles which are on the same base are also in the same parallels. [I. 39]
2
Therefore DE is parallel to BC.
2
Therefore etc. Q. E. D.
PROPOSITION 3.
3
If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.
3
Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC.
3
For let CE be drawn through C parallel to DA, and let BA be carried through and meet it at E.
3
Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29]
3
But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE.
3
Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29]
