Kitap 6
1
Therefore etc. Q. E. D. Under the same height. The Greek text has under the same height AC, with a figure in which the side AC common to the two triangles is perpendicular to the base and is therefore itself the height. But, even if the two triangles are placed contiguously so as to have a common side AC, it is quite gratuitous to require it to be perpendicular to the base. Theon, on this occasion making an improvement, altered to which are (ὅντα) under the same height, (namely) the perpendicular drawn from A to BD. I have ventured to alter so far as to omit AC and to draw the figure in the usual way. ABC, AGB, AHG. Euclid, indifferent to exact order, writes AHG, AGB, ABC. Since then it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD. Here again words have to be supplied in translating the extremely terse Greek ἐπεὶ οὐν ὲδείχθη, ὡς μὲν ἡ βάσις ΒΓ πρὸς τὴν ΓΔ, οὔτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, literally since was proved, as the base BC to CD, so the triangle ABC to the triangle ACD. Cf. note on V. 16, p. 165.
PROPOSITION 2.
2
If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle.
2
For let DE be drawn parallel to BC, one of the sides of the triangle ABC; I say that, as BD is to DA, so is CE to EA.
2
For let BE, CD be joined.
2
Therefore the triangle BDE is equal to the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38]
2
And the triangle ADE is another area.
2
But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.
2
But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1]
