Chapter 7.5 — Exercise 7.4 — Triangle Inequalities

The Two Governing Inequalities

• If two sides of a triangle are unequal, the angle opposite the longer side is the larger angle — and the angle opposite the shorter side is the smaller angle.
• The side opposite a triangle's larger angle is itself the longer side.
• The sum of any two sides of a triangle is always greater than the third side.
• The difference between any two sides of a triangle is always less than the third side.

These four statements are really two ideas viewed from opposite directions each: the first pair links side length to opposite angle size (bigger side ↔ bigger angle), and the second pair links any two sides to the third (their sum is too generous to skip the triangle closed, their difference too stingy to reach all the way around). Unlike the four congruence rules, none of these four statements ever concludes that two triangles match each other — each one instead constrains a single triangle's own sides and angles against one another, which is why every proof in this exercise works with just one triangle (or occasionally two triangles compared side by side) rather than hunting for a congruent pair the way every earlier exercise in this chapter did.

Why a Hypotenuse Is Always the Longest Side

In right triangle ABC with ∠C = 90° and AB the hypotenuse, ∠A + ∠B = 90°, which forces both ∠A and ∠B to be strictly less than 90° individually. Since ∠C is the largest of the three angles, the side opposite it — AB, the hypotenuse — must be the longest side of the triangle: ∠C > ∠A gives AB > BC, and ∠C > ∠B gives AB > AC. This is the reason the hypotenuse is singled out by name in the RHS rule from the previous exercise: it isn't just any side of a right triangle, it's provably the longest one every single time, in every right triangle without exception. This result also explains why RHS only ever needs to check the hypotenuse against a hypotenuse and never risks confusing it with one of the shorter legs — the hypotenuse is always identifiable on sight as the side facing the right angle, and this proof confirms that side is also always the longest, so the two ways of picking it out (facing the 90° angle, or simply the longest side present) agree with each other in every right triangle drawn.

Chaining Two Exterior-Angle Arguments

Sides AB and AC of triangle ABC are extended to points P and Q, and ∠PBC < ∠QCB is given; the goal is to show AC > AB. Since ∠PBC and ∠QCB are exterior angles, ∠PBC = ∠A + ∠ACB and ∠QCB = ∠A + ∠ABC (each exterior angle equals the sum of the two remote interior angles). Substituting, ∠A + ∠ACB < ∠A + ∠ABC, and cancelling ∠A from both sides gives ∠ACB < ∠ABC. The side opposite the smaller angle is the smaller side, so AB < AC — meaning AC > AB, exactly as required.

A companion problem asks the reverse question with a different figure entirely: two triangles OAB and OCD share a vertex O, with ∠B < ∠A in the first and ∠C < ∠D in the second. The smaller-angle-smaller-side rule applied to each triangle separately gives OA < OB and OD < OC; adding these two inequalities gives OA + OD < OB + OC, which is exactly OA + OD compared against OB + OC along the same straight lines — and since OA + OD is the length AD while OB + OC is the length BC, the conclusion AD < BC falls out directly from simply adding two smaller inequalities together. Neither triangle alone contains enough information to reach AD < BC — OAB only ever talks about OA and OB, and OCD only ever talks about OC and OD — so the real insight in this problem is recognising that the two separate inequalities can be added straight down the middle, matching left side to left side and right side to right side, precisely because O sits on both of the straight lines AD and BC at once.

A Quadrilateral's Smallest and Longest Sides

Quadrilateral ABCD has AB as its smallest side and CD as its longest side; joining the two diagonals AC and BD splits the quadrilateral into four small triangles, and comparing angles across those triangles two at a time shows ∠B > ∠D and ∠A > ∠C. In triangle BCD, CD > BC forces the angle opposite CD to exceed the angle opposite BC; in triangle ABD, AD > AB forces a similar inequality — and adding the two relevant angle pairs together each time reconstructs the full angle at B versus at D, then separately at A versus at C. The general lesson holds regardless of the specific quadrilateral: the vertices touching the shortest side end up with the two largest angles, and the vertices touching the longest side end up on the smaller side of each comparison. Drawing both diagonals rather than just one is what makes the proof possible at all: a single diagonal only splits the quadrilateral into two triangles and only exposes one of the two required angle comparisons, but both diagonals together carve out four overlapping triangles, each one isolating a different pair of the quadrilateral's four sides and letting every needed inequality be read off directly from the basic side-versus-angle rule at the top of this lesson.

An Angle Bisector Inside an Unequal Triangle

PR > PQ in triangle PQR, and PS bisects ∠QPR, meeting QR at S; the goal is to prove ∠PSR > ∠PSQ. Since PR > PQ, the angle opposite the longer side is larger: ∠Q > ∠R. Because PS bisects ∠QPR, ∠QPS = ∠RPS exactly. Adding ∠Q > ∠R to this equality gives ∠Q + ∠QPS > ∠R + ∠RPS — and by the exterior-angle property, ∠PSR (exterior to triangle PQS at S) equals ∠Q + ∠QPS, while ∠PSQ (exterior to triangle PSR at S) equals ∠R + ∠RPS. So ∠PSR > ∠PSQ follows directly from the inequality just built. What makes this proof work is that the bisector condition ∠QPS = ∠RPS contributes nothing directional on its own — it's a pure equality, adding the same amount to both sides of the inequality without tipping it either way — so the entire direction of the final result, ∠PSR being the larger of the two angles, traces back entirely to the one genuine inequality given at the start, PR > PQ.

Counting Triangles From Two Given Sides

Both rows lean on exactly the same pair of governing inequalities from the top of this lesson: the difference of two known sides is the strict lower bound on the third side, and their sum is the strict upper bound. The 4 cm/6 cm case has a genuinely wide window, 2 to 10, so seven whole-number lengths fit inside it comfortably. The 5 cm/8 cm/1 cm attempt fails for a very physical reason underneath the algebra: 5 cm and 1 cm together can't even reach the full 8 cm needed to close the triangle back up to its third vertex, since 5 + 1 = 6, which is less than 8 — so no triangle exists no matter how the three sides are arranged, regardless of the angle chosen between them.

The Chapter's Straight-Line and Curved-Shape Neighbors

Triangles built the four congruence rules and these inequality relationships from nothing more than lines and angles meeting at points. Chapter 8, Quadrilaterals extends the same kind of reasoning to four-sided figures next, reusing congruent triangles formed by a quadrilateral's diagonals — exactly the technique already put to work above on the smallest-and-longest-side problem — as one of its central proof tools.