第6卷命题 4 · 等角三角形对应边成比例

Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. For let BC be placed in a straight line with CE. Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet.

设三角形ABC和DCE等角，角ABC等于角DCE，角BAC等于角CDE，角ACB等于角CED。将BC与CE置于同一直线上。

[I. Post. 5] Let them be produced and meet at F. Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28] Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE.

由于角ABC与角ACB小于两直角，且角ACB等于角DEC，故角ABC与角DEC小于两直角，因此延长BA和ED必相交于F。

[I. 28] Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34] And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2] But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE.

由角DCE等于角ABC得BF平行于CD；由角ACB等于角DEC得AC平行于FE。因此FACD为平行四边形，故FA等于DC，AC等于FD。

[V. 16] Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2] But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16] Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE.

由AC平行于FE得BA比AF等于BC比CE，代入AF等于CD得BA比CD等于BC比CE；同理得BC比CE等于AC比DE。综合得BA比AC等于CD比DE。