PROPOSITION 21.¶
On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle;
I say that BD, DC are less than the remaining two sides of the triangle BA, AC, but contain an angle BDC greater than the angle BAC.
For let BD be drawn through to E. Then, since in any triangle two sides are greater than the remaining one, [I. 20]
Again, since, in the triangle CED,
But BA, AC were proved greater than BE, EC;
Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16]
therefore, in the triangle CDE,
For the same reason, moreover, in the triangle ABE also,
But the angle BDC was proved greater than the angle CEB;
See Commentary on Proposition I.21.