第6卷命题 2 · 三角形平行线截比例定理

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. For let BE, CD be joined. 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] And the triangle ADE is another area.

连接BE和CD，则三角形BDE与三角形CDE等积，因同底DE且平行线DE、BC间。

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. 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] For the same reason also, as the triangle CDE is to ADE, so is CE to EA. Therefore also, as BD is to DA, so is CE to EA.

三角形BDE与ADE的比等于BD与DA的比，因同高（从E到AB的垂线）；同理，三角形CDE与ADE的比等于CE与EA的比。

[V. 11] 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. I say that DE is parallel to BC. 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] Therefore each of the triangles BDE, CDE has the same ratio to ADE.

由等量比传递，得BD比DA等于CE比EA。

Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE. But equal triangles which are on the same base are also in the same parallels. [I. 39] Therefore DE is parallel to BC.

反之，若BD比DA等于CE比EA，则三角形BDE与ADE的比等于三角形CDE与ADE的比，故两三角形等积，同底DE，因此DE平行于BC。