If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.
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Just a quick translation here – If you start with a triangle, and from its base construct a new triangle whose third vertex is located within the original triangle, then the two new sides will be less than the original two sides but the angle contained between them will be greater.
We will start with a triangle ABC, and construct a triangle BCD on the base BC such that D is within the triangle.
We will prove that the two sides BD and DC are less than BA and AC, and that the angle BDC is greater than the angle BAC.
Let BD be drawn through to a point E that lies on AC.
In triangle ABE, Proposition 20 (two sides of a triangle together are greater than the remaining side) tells us that sides AB and AE together are greater than BE.
If we add EC to each, we have that AB and AC (AE & EC) together are greater than BE and EC.
We now have that AB and AC greater than two other segments (BE and EC). We need to show that those two sides are also greater than BD and DC.
Looking to triangle CDE, Proposition 20 tells us that the two sides CE and ED together are greater than DC.
If we add BD to each we have that EC and BE (BD & ED) together are greater than DC and BD.
Therefore AB and AC together are greater than BD and DC.
[ AB & AC > BE & EC > BD & DC ]
Now we move on to prove the statement concerning the angles.
(Angle BDC is less than angle BAC)
We will begin by using Proposition 16 (exterior angle of a triangle is greater than either of the opposite interior angles) in triangle CDE.
Angle BDC (exterior angle) is greater than angle CED (an opposite interior angle). Angle CED is equal to angle CEB, so we have angle BDC as greater than angle CEB.
In triangle ABE, the exterior angle CEB is greater than the opposite interior angle BAE (which is equal to angle BAC).
Since angle BDC is greater than angle CEB, which is greater than angle BAC, therefore angle BDC is greater than angle BAC.
Q.E.D.