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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