Triangle inequalities

Pick any three lengths you like — say $5\text{ cm}$ , $9\text{ cm}$ , $3\text{ cm}$ — and try to build a triangle out of them. Sometimes it works. Sometimes the two shorter sides simply cannot reach across to meet the third, and no triangle exists at all, no matter how you tilt the sticks. Triangles are not free-for-all shapes: the lengths of their sides and the sizes of their opposite angles are locked together by a small set of strict rules called triangle inequalities. These rules let you answer geometry questions without a ruler or protractor — which side is longest, whether a triangle can even be drawn, what range an unknown side must fall in, and why the perpendicular from a point to a line is always the shortest path to it. This article builds up the core inequality from a simple picture, applies it to worked ICSE-style problems, and flags the traps that turn a one-mark question into a mistake.

The core insight

In any triangle, sides and their opposite angles rise and fall together — the greater side sits opposite the greater angle — and the sum of any two sides must strictly exceed the third, $a+b>c$ , or no triangle exists.

The core idea

There are four linked results here, all about inequality, not equality — none of them apply if the triangle is equilateral or the two things being compared are equal.

1. Equal sides pair with equal angles. In $\triangle ABC$ , $AB = AC$ if and only if $\angle B = \angle C$ . This is the familiar isosceles-triangle fact, and it is the equality case that the next result generalizes.

2. Bigger side pairs with bigger opposite angle. If two sides of a triangle are unequal, the angle opposite the longer side is the larger angle, and conversely the side opposite the larger angle is the longer side. For $\triangle ABC$ :

AB > AC \iff \angle C > \angle B

where: angle $C$ is the angle that does not touch side $AB$ (so it is "opposite" $AB$ ), and angle $B$ is opposite $AC$ . Side and angle move together — you never get a triangle where the longest side faces the smallest angle.

3. Sum of two sides beats the third. For any triangle with sides $a$ , $b$ , $c$ :

a + b > c, \qquad b + c > a, \qquad c + a > b

where: $a$ , $b$ , $c$ are the three side lengths of the triangle, in any consistent unit. Equivalently, packing all three into one statement, each side is bounded between the sum and the difference of the other two:

|b - c| < a < b + c

where: $|b-c|$ is the absolute difference of the two known sides and $b+c$ is their sum, so this single inequality gives both the lower and upper bound on the third side $a$ at once.

4. Perpendicular is shortest. Of all line segments drawn from a point outside a line to that line, the perpendicular segment is the shortest. Any other segment forms the hypotenuse of a right triangle with the perpendicular as one leg, and a hypotenuse is always longer than either leg.

Figure

Two triangles side by side, one with two sides that reach to close the third side and one where two short sides fail to reach across a longer base, illustrating when the triangle inequality does and does not hold

Derivation AWhy the perpendicular from a point to a line is the shortest segment to that line

We want to show that among every line segment drawn from an external point $P$ to a fixed line $\ell$ , the one that meets $\ell$ at a right angle has strictly the smallest length, so that any slanted segment from $P$ to $\ell$ is always longer than the perpendicular and never equal or shorter to it, using only the fact that a right triangle's hypotenuse exceeds either of its legs.

Step 1. Drop the perpendicular from $P$ to $\ell$ , meeting $\ell$ at foot $M$ , so $PM \perp \ell$ and $PM$ is the candidate shortest segment. Let $N$ be any other point on $\ell$ , $N \neq M$ , so $PN$ is a "slanted" segment we want to compare with $PM$ .

Step 2. The points $P$ , $M$ , $N$ form a triangle $\triangle PMN$ , and because $PM \perp \ell$ and $MN$ lies along $\ell$ , the angle at $M$ inside this triangle is $\angle PMN = 90^\circ$ . So $\triangle PMN$ is a right triangle with the right angle at $M$ .

Step 3. In a right triangle, the side opposite the right angle is the hypotenuse, and by result 2 above (bigger angle opposite bigger side), the side opposite the largest angle is the longest side of the triangle. Since $90^\circ$ is the largest possible angle sum contribution here (the other two angles of $\triangle PMN$ must sum to $90^\circ$ , so each is less than $90^\circ$ ), the hypotenuse $PN$ is strictly longer than either leg, in particular strictly longer than leg $PM$ .

Step 4. Therefore $PN > PM$ for every point $N$ on $\ell$ other than $M$ . Since $N$ was an arbitrary point on the line other than the foot of the perpendicular, this holds for every slanted segment, so $PM$ — the perpendicular — is shorter than all of them, proving it is the unique shortest distance from $P$ to $\ell$ .

Worked example

In $\triangle PQR$ , $\angle Q = 70^\circ$ and $\angle R = 55^\circ$ , with $PQ = 6\text{ cm}$ . Find $\angle P$ , decide which side of the triangle is longest, and find $QR$ — all without Pythagoras or the sine rule. Then check whether $PQ = 6\text{ cm}$ , $QR = 6\text{ cm}$ , and a third side $PR$ of length $13\text{ cm}$ could all belong to the same triangle.

Step 1. Use the angle-sum property of a triangle to find the missing angle:

\angle P = 180^\circ - 70^\circ - 55^\circ = 55^\circ

So the three angles are $\angle P = 55^\circ$ , $\angle Q = 70^\circ$ , $\angle R = 55^\circ$ .

Step 2. Compare the angles: $\angle Q = 70^\circ$ is strictly the largest, while $\angle P = \angle R = 55^\circ$ are equal to each other. By result 1 (equal angles pair with equal opposite sides), since $\angle P = \angle R$ , the sides opposite them are equal. The side opposite $\angle P$ is $QR$ , and the side opposite $\angle R$ is $PQ$ , so

QR = PQ = 6\text{ cm}

Step 3. By result 2 (bigger angle pairs with bigger opposite side), $\angle Q$ is the largest angle, and the side opposite it is $PR$ . So $PR$ is the longest side of $\triangle PQR$ — longer than $6\text{ cm}$ , though its exact length needs more than angle comparison to pin down.

Step 4. Now check the proposed triple $PQ = 6$ , $QR = 6$ , $PR = 13$ against result 3, $a + b > c$ for every pair. Take the two smaller sides: $PQ + QR = 6 + 6 = 12$ , but $12 < 13$ . Since the sum of the two shorter sides does not exceed the third, these three lengths cannot form a triangle — so $PR$ cannot actually equal $13\text{ cm}$ here.

Step 5. To find the true allowed range for $PR$ given $PQ = QR = 6\text{ cm}$ , apply $|b-c| < a < b+c$ with $b = c = 6$ :

|6-6| < PR < 6+6 \implies 0 < PR < 12

Combined with Step 3 (that $PR$ is the longest side, so $PR > 6$ ), the length of $PR$ must satisfy $\boxed{6\text{ cm} < PR < 12\text{ cm}}$ .

Competitive stretch

If instead only $PQ = 5\text{ cm}$ and $QR = 9\text{ cm}$ were known (dropping the equal-angle condition), the same bound $|b-c| < a < b+c$ gives $9 - 5 < PR < 9 + 5$ , i.e. $\boxed{4\text{ cm} < PR < 14\text{ cm}}$ — a wider window than the isosceles case above, since fixing two sides equal (via equal angles) also pins the third side into a tighter, symmetric range.

Common traps

Writing " $\geq$ " instead of " $>$ ". The inequality is strict: if $a + b = c$ exactly, the three "vertices" become collinear — a flattened, zero-area figure, not a genuine triangle.
Matching the wrong side to the wrong angle. The side opposite an angle is the one that does not touch that angle's vertex; swapping this pairing flips the inequality and gives the wrong "longest side."
Assuming any slanted segment beats the perpendicular. Only the perpendicular from an external point to a line is guaranteed shortest; every other segment to that line is strictly longer, never equal.
Skipping the consistency check. Before trusting three given lengths as a triangle, always test $a+b>c$ for the two smaller sides against the largest — this is the one check that instantly rules out impossible data.

Ask the tutor: Want to test whether a tricky set of three lengths can form a triangle, or practise ordering sides and angles in a labelled figure? Ask the AI tutor to work through a few more triangle inequality problems with you.

Why is the triangle inequality strict, not allowing a + b = c?

When $a + b = c$ exactly, imagine the two shorter sides swinging to meet — at that exact length they lie flat along the third side instead of rising to meet at a peak. The "triangle" has zero height and zero area: it is really just a straight line segment covered twice. Geometry only calls something a triangle if the three vertices are non-collinear, so the sum must be strictly greater, not equal.

Do I need to prove these results in ICSE exams, or just use them?

At Class 9 ICSE level, the syllabus explicitly treats angle-side inequalities and the sum-of-two-sides rule as results to apply, not to prove — proofs are reserved for later classes. You are expected to quote the rule, plug in the given angles or sides, and reach a numeric or comparative conclusion, as in the worked example above. The one exception worth knowing informally is the perpendicular-shortest-distance result, whose reasoning (hypotenuse beats a leg) is short enough to be asked as a "explain why" question.

How is |b - c| < a < b + c different from just writing three separate inequalities?

They say the same thing, just packaged differently. Writing $a+b>c$ , $b+c>a$ , $c+a>b$ as three separate statements is exact and works for any labelling. Writing $|b-c| < a < b+c$ isolates one particular side, $a$ , and gives both its lower and upper bound in a single line — which is exactly the form you need when a question asks "what is the range of the third side," since you only care about bounding the unknown one, not re-deriving all three inequalities from scratch.

Triangle inequalities

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads