Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD, DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC.
Produce BD to meet AC at E.
Because two sides of a triangle are greater than the third side, the two sides BA, AE of the triangle ABE are greater than the side BE. [I. 20.
To each of these add EC.
Therefore BA, AC are greater than BE, EC.
Again; the two sides CE, ED of the triangle CED are greater than the third side CD. [I. 20.
To each of these add DB.
Therefore CE, EB are greater than CD, DB.
But it has been shewn that BA, AC are greater than BE, EC;
much more then are BA, AC greater than BD, DC.
Again, because the exterior angle of any triangle is greater than the interior opposite angle, the exterior angle BDC of the triangle CDE is greater than the angle CED. [I. 16.
For the same reason, the exterior angle CEB of the triangle ABE is greater than the angle BAE.
But it has been shewn that the angle BDC is greater than the angle CEB;
much more then is the angle BDC greater than the angle BAC.
Wherefore, if from the ends &c. q.e.d.
