第6卷命题 3 · 三角形内角平分线定理

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC. For let CE be drawn through C parallel to DA, and let BA be carried through and meet it at E. Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29] But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE.

过C作CE平行于AD，交BA延长线于E。

Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29] But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6] And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE.

由平行线性质，∠ACE=∠CAD，又∠CAD=∠BAD，故∠BAD=∠ACE；同理∠BAD=∠AEC，所以∠ACE=∠AEC，得AE=AC。

But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC. Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A. D. For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE : for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE.

在△BCE中，AD平行于EC，由比例定理得BD:DC=BA:AE，代入AE=AC得BD:DC=BA:AC。

[V. 11] Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5] But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.] therefore the angle BAD is also equal to the angle CAD. Therefore the angle BAC has been bisected by the straight line AD.

反之，设BD:DC=BA:AC，作同样辅助线，由比例得BA:AC=BA:AE，故AC=AE，则∠AEC=∠ACE，进而∠BAD=∠CAD，即AD平分∠BAC。