elem.6.7
若两三角形有一角相等，且该角两边与另一角两边成比例，而其余两角均小于直角或均不小于直角，则这两三角形等角，且对应边成比例。
三角形 ABC 与 DEF：∠BAC=∠EDF，AB:BC=DE:EF；反证假设 ∠ABC 与 ∠DEF 不等，在 AB 上构造 ∠ABG=∠DEF 得辅助点 G，由角与边关系推出矛盾。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F. For, if the angle ABC is unequal to the angle DEF, one of them is greater. Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23] Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32] Therefore the triangle ABG is equiangular with the triangle DEF.
假设角ABC大于角DEF，在AB上作角ABG等于角DEF，则三角形ABG与DEF等角，故AB比BG等于DE比EF。
Therefore, as AB is to BG, so is DE to EF [VI. 4] But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9] so that the angle at C is also equal to the angle BGC. [I. 5] But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13] And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle. But it is by hypothesis less than a right angle : which is absurd. Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it.
由假设AB比BC等于DE比EF，得AB比BC等于AB比BG，因此BC等于BG，从而角C等于角BGC。
But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF. But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF. For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5] But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.
若角C小于直角，则角BGC也小于直角，其邻角AGB大于直角，但角AGB等于角F，故角F大于直角，与假设矛盾。
Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17] Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF.
若角C不小于直角，则角BGC也不小于直角，三角形BGC中两角不小于两直角，与三角形内角和定理矛盾。因此角ABC等于角DEF，从而两三角形等角。