**Triangle Inequality** **
***Printout*

*Proof is
the idol before whom the pure mathematician tortures himself.
—*

** **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.//

© Copyright 2005, 2006 - Timothy Peil |