Three sides, three angles, the sturdiest frame there is — and the rules they must obey.
Point 2 of 6 in this lesson: 15.1.2 How the three sides relate — the triangle inequality
Three straight pieces, joined end to end, fence off the simplest flat shape there is — and the strongest. Push on a four-sided frame and it racks over into a leaning parallelogram; push on a triangle and it will not budge. That stubborn rigidity is why bridges, cranes, and roof trusses are stitched together out of triangles, and it is the secret engine of this entire stage. Before we can prove things with triangles, let's properly meet one. We'll name its three sides and three angles, then collect the small handful of rules they can never break — how the sides must relate, why the angles always total 180°, how to sort triangles into families, and the three special lines you can draw inside any one of them.
Start with three points that are not all on one line — we call such points non‑collinear. Join them in pairs with three straight segments, and the segments close up into a triangle. The three corner points are the vertices (one vertex, two vertices); the three segments are the sides.
We name the triangle by its vertices: △ABC. Each side is named with the lowercase letter of the vertex it is opposite — the side that doesn't touch that corner:
If the three points were on one line, the "triangle" would flatten into a segment with no inside — that's why we insist the points are non‑collinear. And once the three sides are fixed, a triangle cannot flex: its shape is locked. A four-sided frame can sag; a triangle stays put. That single fact — rigidity — is what makes the triangle the building block of geometry.
Not every trio of lengths can close into a triangle. Take sticks of length 1, 2, and 5: lay the long one down, and the two short ones together only reach a length of 3 — they can't stretch far enough to meet. The triangle never closes.
The rule behind this is the triangle inequality: any two sides together must be longer than the third. For △ABC all three of these must hold at once:
a + b > c, b + c > a, c + a > b.
The straight segment between two points is the shortest route there is. Walking from B to C directly costs distance a; detouring through A costs c + b. A detour can't be shorter than the straight path, so c + b > a. The same argument at the other two vertices gives the other two inequalities. If two sides only equal the third (2 + 3 = 5), the triangle collapses flat onto the long side.
Here is the most useful fact about triangles. Tear the three corners off a paper triangle and lay them snugly together, tip to tip. They pave a perfectly straight line — a straight angle. So the three interior angles of any triangle, no matter its shape, total a half‑turn:
∠A + ∠B + ∠C = 180°.
Through vertex A, draw a line parallel to side BC. Side AB is a transversal, so the angle it makes on the far side equals ∠B (alternate interior angles — see Properties of Parallel Lines). Likewise side AC peels off an angle equal to ∠C. Those two angles plus ∠A sit on one straight line through A, so they add to 180° — and therefore ∠A + ∠B + ∠C = 180°.
Triangles come in families, sorted two different ways — by their sides, and by their angles.
A triangle can have at most one right or obtuse angle. Why? If two angles were each 90° or more, they would already use up the whole 180° budget (or overspend it), leaving nothing for the third angle.
| acute | right | obtuse | |
|---|---|---|---|
| scalene | 5, 6, 7 | 3, 4, 5 | 4, 5, 7 |
| isosceles | 5, 5, 6 | 1, 1, √2 | 5, 5, 9 |
| equilateral | 6, 6, 6 | — | — |
An equilateral triangle is always acute (every angle is 60°), so two cells are impossible.
Extend one side of the triangle past a vertex. The angle that opens up outside the triangle, between that extension and the next side, is an exterior angle. It sits right next to one interior angle — they're a linear pair, so they add to 180°.
The beautiful part is what the exterior angle equals on the inside. The Exterior Angle Theorem says:
an exterior angle = the sum of the two remote (non-adjacent) interior angles.
Call the interior angle at C the value ∠C. The exterior angle beside it is 180° − ∠C (linear pair). But the angle sum says ∠C = 180° − (∠A + ∠B), so 180° − ∠C = ∠A + ∠B. The exterior angle equals the other two interior angles added together. As a bonus, the three exterior angles (one at each vertex) always sum to 360° — a full turn.
From any vertex you can draw three different special segments down into the triangle. Each is a kind of cevian — a segment from a vertex to the opposite side. There are three of each, but let's draw one of each from the same vertex A:
In most triangles the altitude, median, and bisector from one vertex are three separate lines landing at three different spots on the opposite side. They only merge into one segment in a special case — when the two sides from that vertex are equal (an isosceles triangle). We'll see exactly that collapse in Isosceles Triangles.
A triangle joins three non‑collinear points; sides are named a, b, c for the opposite vertices, and the shape is rigid.
Triangle inequality: any two sides beat the third — a+b>c, and the other two ways round. Otherwise the triangle can't close.
Angle sum: ∠A+∠B+∠C = 180°, always.
Families: by sides — scalene / isosceles / equilateral; by angles — acute / right / obtuse (at most one right or obtuse angle).
Exterior angle = the two remote interior angles added; the three exterior angles total 360°.
Cevians: the altitude (height, ⊥ to the opposite side), median (to the midpoint), and angle bisector — three different lines in general.
Can three sticks of length 3, 4, and 5 form a triangle? Can 1, 2, and 5?
3, 4, 5: the two shorter sides give 3 + 4 = 7 > 5, so yes — they close. For 1, 2, 5: 1 + 2 = 3, which is not greater than 5, so no — too short to close.
Two angles of a triangle are 50° and 60°. Find the third angle.
The three add to 180°, so the third is 180° − 50° − 60° = 70°.
Classify the triangle with sides 5, 5, 8 by its sides, and the triangle with angles 40°, 40°, 100° by its angles.
Sides 5, 5, 8 have two equal sides → isosceles. Angles 40°, 40°, 100° include one angle over 90° → obtuse. (And they sum to 180° — good.)
An exterior angle of a triangle has the two remote interior angles 50° and 60°. How big is the exterior angle?
By the Exterior Angle Theorem it equals the two remote interiors added: 50° + 60° = 110°. (Check: the interior angle beside it is 70°, and 110° + 70° = 180°.)
The three angles of a triangle are x, 2x, and 3x. Solve for x and name the triangle by its angles.
x + 2x + 3x = 6x = 180°, so x = 30°. The angles are 30°, 60°, 90° — a right triangle.
Explain why a triangle cannot have two right angles.
Two right angles already total 90° + 90° = 180°, using up the whole angle sum. That would leave 0° for the third angle — but every triangle has three genuine angles, so it's impossible. (The same reasoning blocks two obtuse angles.)
Six questions to lock it in. Tap the answer you think is right.
This lesson is the doorway to school geometry: the triangle as the first shape we can truly reason about. Three ideas carry the most weight — the triangle inequality (which three lengths actually close), the 180° angle sum (with the parallel-line argument behind it), and the classification by sides and angles. Everything later in the stage — congruence, isosceles symmetry, the Pythagorean theorem — leans on these.
Two misconceptions are worth catching early. The first is believing that any three lengths make a triangle; the can-these-close widget exists to make the failure visible — when the two short sticks can't reach across the long one, the triangle simply won't form. The second is blurring the three cevians together: students often say "the height goes to the midpoint" or "the median is perpendicular." In a general triangle the altitude, median, and angle bisector are three distinct lines; they only coincide in the isosceles case (15.5). Drawing all three from one vertex, in three colors, helps the distinction stick.
Aligned to CCSS 7.G.A.2 (conditions that determine a triangle), 8.G.A.5 (angle sum and the exterior angle), 4.G.A.2 (classify by sides and angles), and HS G‑CO.C.10 (prove theorems about triangles, including the angle-sum and exterior-angle results).