Kitap 6
1
Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF, therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]
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.
