If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.
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.
But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle
BAD is also equal to the angle ACE.
Again, since the straight line BAE falls upon the parallels AD, EC, the
exterior angle BAD is equal to the interior angle AEC.
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.
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.
But AE is equal to AC; 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 AD.
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:
therefore also, as BA is to AC, so is BA to AE.
Therefore AC is equal to AE,
so that the angle AEC is also equal to the angle ACE.
But the angle AEC is equal to the exterior angle BAD,
and the angle ACE is equal to the alternate angle CAD;
therefore the angle BAD is also equal to the angle CAD.
Therefore the angle BAC has been bisected by the straight line AD.
Therefore etc.
Q.E.D