3.7 Inequalities in a Single Triangle
We have spent a great deal of time so far dealing with equalities and congruences. We conclude this chapter with three theorems having to do with inequalities in a triangle.
- Theorem 3.18: If two sides of a triangle are not congruent, then the angles opposite those sides are not congruent, and the larger angle is opposite the longer side.
Given: In $\triangle
ABC$, $BC>AC$. Prove: $m\angle CAB > m\angle B$ Construction: Let $D$ be the point on $\overline{CB}$ such that $\overline{CD}\cong \overline{AC}$. |
Prove: $\triangle ACD$ is isosceles, byt construction, so $m\angle CAD = m\angle CDA$. Since $D$ is in teh interior of $\angle CAB$, $m\angle CAD + m\angle DAB = m\angle CAB$. And since $m\angle DAB$ is positive, $m\angle CAB > m\angle CAD$. By the Exterior Angle Inequality, $m\angle CDA>m\angle B$. Putting these all together, we have $m\angle CAB > m\angle CAD = m\angle CDA > m \angle B$.
Example:
In $\triangle XYZ$, $XY=10,\;YZ=7$, and $XZ=12$. Arrange
the angles in order of increasing size. Solution: The longest side of the triangle $\overline{XZ}$, so the angle opposite that side is larger than either of teh other two angles. The shortest side is $\overline{YZ}$, so the angle opposite that side is the smallest angle. Hence, $m\angle X< m\angle Z< m\angle Y$. |
We now prove the converse of Theorem 3.15.
- Theorem 3.19:
If two angles of a triangle are not congruent, then the sides opposite
those angles are not congruent, and the longer side is opposite the
larger angle.
Given: In $\triangle
ABC$, $m\angle A> m\angle B$. Prove: $BC>AC$ |
Proof: We are given that $m\angle A>m\angle B$. When we compare the lengths of the opposite sides $\overline{BC}$ and $\overline{AC}$, there are three possibilities, one of which must be true: $$BC=AC,\;\;BC<AC,\;\;BC>AC.$$
$BC=AC$ is impossible, because if it were the case, then $\angle A\cong \angle B$ by the Isosceles Triangle Theorem, and we are given that $m\angle A > m\angle B$.
$BC<AC$ is also impossible, because if it were the case, then (by the last theorem) $m\angle A<m\angle B$, which also contradicts the given.
That leaves $BC>AC$ by a process of elimination.
Example: In the figure to the right (which is not drawn to scale), with angle measures as marked, which is the longest segment? |
Solution: In $\triangle ABD$, we can arrange the sides from smallest to largest: $$BD<AD<AB.$$
In $\triangle BCD$, we can also arrange the sides from smallest to largest: $$CD<BC<BD.$$
Since the shortest side of $\triangle ABD$is the longest side of $\triangle BCD$, we can string these inequalities together to get $$CD<BC<BD<AD<AB.$$
Hence the longest segment in the figure is $\overline{AB}$.
The significance of this example is that it emphasizes the fact that Theorem 3.15 and Theorem 3.16 apply only to a single triangle. You can’t say that the longest segment in a diagram is opposite the largest angle in the diagram.
We conclude with a theorem that has many applications. It looks obvious, but the proof is not all that easy.
- Theorem 3.20:
The Triangle Inequality
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Given: $\triangle ABC$ Prove: $AB+BC>AC$ Construction: Exend $\overline{BC}$ to the point $D$ such that $\overline{BD}\cong\overline{BA}$. |
Proof: Since $\overline{BD}\cong\overline{BA}$, we know that $m\angle D=m\angle DAB$.
Since $B$ is in the interior of $\angle DAC$, we have $m\angle DAC=m\angle BAC+m\angle DAB$.
As in the proof of the first theorem of this section, since $m\angle BAC$ is positive, we have $m\angle DAC>m\angle DAB$.
That means that $m\angle DAC>m\angle D$.
Then, using Theorem 3.16, since $\angle DAC$ and $\angle D$ are not congruent, the sides opposite them (in $\triangle DAC$) are not congruent, with the longer side lying opposite the larger angle. Hence $DC>AC$. But $DC=DB+BC=AB+BC$. Therefore, $AB+BC>AC$.
We add an important note about this proof. We didn’t bother to prove the other two cases: $AB+AC>BC$ and $BC+AC>AB$. Those proofs are done in exactly the same way as this one --- by extending a side to create an isosceles triangle.