therefore AC is not less than AB.
And it has been shewn that AC is not equal to AB.
Therefore AC is greater than AB.
Wherefore, the greater angle &c. q.e.d.
PROPOSITION 20. THEOREM.
Any two sides of a triangle are together greater than the third side.
Let ABC be a triangle: any two sides of it are together greater than the third side; namely, BA, AC greater than BC; and AB, BC greater than AC; and BC, CA greater than AB.
Produce BA to D,
making AD equal to AC, [I. 3.
and join DC.
Then, because AD is equal to AC, [Construction.
the angle ADC is equal to the angle ACD. [I. 5.
But the angle BCD is greater than the angle ACD. [Ax. 9.
Therefore the angle BCD is greater than the angle BDC.
And because the angle BCD of the triangle BCD is greater than its angle BDC, and that the greater angle is subtended by the greater side; [I. 19.
therefore the side BD is greater than the side BC.
But BD is equal to BA and AC.
Therefore BA, AC are greater than BC.
In the same manner it may be shewn that AB, BC are greater than AC, and BC, CA greater than AB.
Wherefore, any two sides &c. q.e.d.
PROPOSITION 21. THEOREM.
If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.