Triangle Inequality
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Proof is
the idol before whom the pure mathematician tortures himself.
—Sir Arthur Eddington (1882–1944)
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.//
Triangle Inequality. In a neutral geometry, the length of one side of a triangle is strictly less than the sum of the lengths of the other two sides.
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.//
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