On this page, we prove the Triangle Inequality based on neutral geometry results from Chapter 2.
Lemma. In a neutral geometry, if one angle is greater in measure than another angle of a triangle, then the opposite side of the greater angle is longer than the opposite side of the lessor angle. Furthermore, the longest side of a triangle is opposite the angle of greatest measure.
Proof. Given . Assume Let A' be on ray such that Then One of the following three is true: B is between C and A'; A' = B; or A' is between C and B.
Case 1. Assume B is between C and A'. Then and by the
Exterior Angle Theorem Thus Hence ,
which contradicts the initial assumption. Hence B is not between C and A'.
Case 2. Assume A' = B. Then Hence , which contradicts the initial assumption.
By Cases 1 and 2, we must have A' between C and B. Hence, BC = BA' + A'C > A'C = AC. The cases for the other pairs of sides may be proved similarly. Therefore, the longest side of a triangle is opposite the angle of greatest measure.//
Proof. Given . We prove one case here; the other cases are similar. Let D be on ray such that C is between B and D and Since is an isosceles triangle, Since C is between B and D and , , by the Angle Addition Postulate. Hence By the Lemma, BD > BA. Therefore, since C is between B and D, AC + CB = DC + CB = DB > BA. Since the proofs of the other two cases are similar, the length of one side of a triangle is strictly less than the sum of the lengths of the other two sides.//
© Copyright 2005, 2006 - Timothy Peil