Inequalities in a Triangle

So far the chapter has been about equalities: matching sides, matching angles, congruent triangles. This lesson is about inequalities , what happens when sides or angles are not equal. The two key facts: (1) the side opposite the larger angle is longer, and (2) the sum of any two sides of a triangle exceeds the third. Both are essential geometric facts you will use forever.

Definitions

In $\triangle ABC$ :

The side $a = BC$ is opposite the angle $\angle A$ .
The side $b = CA$ is opposite $\angle B$ .
The side $c = AB$ is opposite $\angle C$ .

A non-isosceles triangle has three different side lengths and three different angles.

The two theorems

Theorem 1 (Side opposite larger angle is longer). In a triangle, the side opposite the larger angle is longer than the side opposite the smaller angle. Conversely, the angle opposite the longer side is larger than the angle opposite the shorter side.

Theorem 2 (Triangle inequality). In any triangle, the sum of the lengths of any two sides is greater than the length of the third side.

Proof sketches

Theorem 1. Suppose $\angle B > \angle C$ in $\triangle ABC$ . We claim $AC > AB$ . Pick a point $D$ on $AC$ such that $AD = AB$ . Then $\triangle ABD$ is isosceles ( $AB = AD$ ), so $\angle ABD = \angle ADB$ . Now $\angle ADB$ is an exterior angle of $\triangle BDC$ , so by the exterior-angle theorem it is greater than $\angle DCB = \angle C$ . Therefore $\angle ABD > \angle C$ , and so $\angle B$ (which is larger than $\angle ABD$ since $D$ is in the interior) is also greater than $\angle C$ . Working backwards through the construction, this forces $AC = AD + DC > AB$ . (The argument is best followed with a careful figure.)

Theorem 2. In $\triangle ABC$ , claim $AB + AC > BC$ . Extend $BA$ beyond $A$ to $D$ such that $AD = AC$ . Then $\triangle ADC$ is isosceles, so $\angle ADC = \angle ACD$ . Now $\angle BCD = \angle BCA + \angle ACD > \angle ACD = \angle ADC = \angle BDC$ . By Theorem 1 applied to $\triangle BCD$ , the side opposite the larger angle is longer: $BD > BC$ . But $BD = BA + AD = BA + AC$ , so $BA + AC > BC$ . Q.E.D.

Practical consequences

Triangle inequality in all three forms.

$a + b > c$
$b + c > a$
$c + a > b$

So given three positive lengths $a, b, c$ , you can form a triangle iff all three of these inequalities hold. (Equivalently, the largest of the three is less than the sum of the other two.)

Determining which side is longest. Compute the angles; the longest side is opposite the largest angle.

Determining which angle is largest. Compute the sides; the largest angle is opposite the longest side.

Worked examples

Example 1. Can a triangle have sides $5, 7, 13$ ?

Check: $5 + 7 = 12 < 13$ . The inequality fails , no triangle exists.

Example 2. A triangle has sides $4, 7, x$ . Find the range of $x$ .

Triangle inequality:

$4 + 7 > x \Rightarrow x < 11$ .
$4 + x > 7 \Rightarrow x > 3$ .
$7 + x > 4 \Rightarrow x > -3$ (always true).

So $3 < x < 11$ .

Example 3. In $\triangle ABC$ , $\angle A = 80^\circ$ , $\angle B = 60^\circ$ , $\angle C = 40^\circ$ . Which is the longest side?

$\angle A$ is largest, so $BC$ (opposite $\angle A$ ) is longest.

Example 4. In $\triangle ABC$ , $AB = 5$ , $BC = 8$ , $CA = 6$ . Which is the largest angle?

$BC = 8$ is longest, so $\angle A$ (opposite $BC$ ) is largest.

Example 5. Two sides of a triangle are $7$ and $10$ . What is the smallest possible value of the third side, given it is a positive integer?

Third side $x$ must satisfy $10 - 7 < x < 10 + 7$ , i.e. $3 < x < 17$ . Smallest positive integer in this range: $x = 4$ .

Try it yourself

State the triangle inequality.
Can a triangle have sides $3, 4, 9$ ?
A triangle has sides $5, 12, x$ . Find the range of $x$ .
State which is true: "side opposite the smaller angle is longer".
In $\triangle ABC$ , $\angle A = 90^\circ$ . Which side is longest? Why?
In $\triangle ABC$ , $AB = 5, BC = 7, CA = 9$ . Order the angles by size.
Two sides of a triangle are $4$ and $9$ . What integer values of the third side are possible?
State whether each is a valid triangle: $(2, 2, 3), (1, 1, 2), (5, 12, 13), (8, 8, 8)$ .
Prove: in any triangle, the longest side is opposite the largest angle.
A triangle has angles $30^\circ, 70^\circ, 80^\circ$ . List the sides from shortest to longest (by opposite angle).

Pitfalls / Insight

Triangle inequality is strict. Equality means the three points are collinear and the "triangle" is degenerate.
Compare angles to compare sides, and vice versa. This is the essence of Theorem 1.
Three inequalities, all required. Don't stop after checking one , all three must hold.

Insight. These inequalities, simple as they look, are the geometric form of "straight line is the shortest distance between two points". If you tried to go from $B$ to $C$ via $A$ , you'd cover $BA + AC$ , strictly more than going directly along $BC$ . That intuition is the entire content of the triangle inequality.