Prove that the difference of any two sides of a triangle is less than the third side.

Cut off AD = AB, join BD

2 = 4

Exterior 1 > 4      (exterior angle is greater than each of the interior angles)

Exterior 2 > 3

1 > 3

BC > DC                (side opposite greater angle is greater)

BC > AC - AD        (DC = AC - AD)

BC > AC - AB         (AD = AB)

BC > AC - AB < BC

Similarly, BC - AC < AB and BC - AC < AC

Hence, the difference of any two sides of a triangle is less than the third side.