第6卷命题 6 · 两边成比例且夹角相等则三角形相似

Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE. For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32] Therefore the triangle ABC is equiangular with the triangle DGF.

设三角形ABC和DEF中，角BAC等于角EDF，且BA比AC等于ED比DF。

Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4] But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF.

在直线DF上，于点D和F作角FDG等于角BAC或角EDF，角DFG等于角ACB，则三角形ABC与DGF各角对应相等。

[V. 11] Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF.

由相似三角形对应边成比例，得BA比AC等于GD比DF；结合已知比例，得ED比DF等于GD比DF，故ED等于DG。

But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE. And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF.

由边角边定理得三角形DEF全等于DGF，故角DFG等于角DFE，角DGF等于角DEF；又角DFG等于角ACB，角BAC等于角EDF，因此三角形ABC与DEF各角对应相等。