Topic · Geometry

Triangles

The simplest closed shape in the plane: three points, three segments, three angles. Everything else in classical geometry — polygons, areas, the Pythagorean theorem, trigonometry — is built by gluing triangles together or breaking shapes back down into them.

What you'll leave with

  • A precise definition of a triangle and why "non-collinear" matters.
  • The six standard classifications — three by sides, three by angles — and why they're independent.
  • The angle-sum theorem ($\alpha + \beta + \gamma = 180°$) and a one-glance proof using parallel lines.
  • Five congruence rules (SSS, SAS, ASA, AAS) and the famous one that isn't a rule (SSA).
  • The triangle inequality: why you can't make a triangle out of sides $1$, $2$, and $5$.

1. What a triangle is

Triangle

Three line segments connecting three points that don't all lie on the same line. The three points are the vertices; the segments between them are the sides; and at each vertex, the two sides meeting there form an interior angle.

The "don't all lie on the same line" clause — the technical word is non-collinear — is doing all the work. If your three points sit on a single line, the "segments" between them just trace and retrace that line. You'd get no enclosed region and no angles to speak of. The triangle would degenerate into a flat sliver of nothing.

A triangle with vertices $A$, $B$, $C$ is written $\triangle ABC$. It carries six pieces of data: three side lengths (call them $a$, $b$, $c$, where lowercase $a$ is the side opposite uppercase vertex $A$, and so on) and three angles ($\alpha$, $\beta$, $\gamma$, at vertices $A$, $B$, $C$).

A B C a b c α β γ
Triangle $ABC$ with vertices, sides, and angles labeled by convention.

That naming convention — side $a$ across from vertex $A$ — is universal in geometry and trigonometry. Memorize it once and never untangle "which side is which" again.

2. Classifying by sides

The first way we sort triangles is by counting how many sides have equal length.

  • Equilateral — all three sides are equal. $a = b = c$.
  • Isosceles — at least two sides are equal.
  • Scalene — no two sides are equal. All three are different lengths.

"At least two" rather than "exactly two" matters: under most modern conventions, an equilateral triangle is a special case of an isosceles triangle — it has two equal sides, plus a bonus third. This is the same logic by which a square is a special rectangle.

Equilateral a = b = c Isosceles two sides equal Scalene no sides equal
Tick marks indicate equal sides. The equilateral has three pairs of marks; the isosceles, two.

A handy property: the sides and angles of a triangle pair up. Equal sides force equal opposite angles. An equilateral triangle therefore has three equal angles (each $60°$, by the angle-sum theorem in section 4). An isosceles triangle has the two angles opposite its equal sides equal — called the base angles. A scalene triangle has three different angles.

3. Classifying by angles

The second way we sort triangles is by their largest angle.

  • Acute — every angle is less than $90°$.
  • Right — exactly one angle is $90°$. (You can't have two: their sum would already hit $180°$, leaving nothing for the third.)
  • Obtuse — one angle is greater than $90°$. (Again, only one possible.)
Acute all angles < 90° Right one angle = 90° Obtuse one angle > 90°
The right triangle's $90°$ corner is marked with the customary small square; the obtuse triangle's wide angle with an arc.
Classifications stack

The two classifications — by sides, by angles — are independent. Every triangle has one label from each axis. A right isosceles triangle has a $90°$ angle and two equal legs (with angles $45°$, $45°$, $90°$). An obtuse scalene triangle has one angle past $90°$ and all three sides different. There's no equilateral right or equilateral obtuse, though — every equilateral triangle is acute, with three $60°$ angles.

4. The angle-sum theorem

One of the most-used facts in all of geometry:

$$ \alpha + \beta + \gamma = 180° $$

The three interior angles of any triangle, no matter how stretched or squashed, always sum to exactly $180°$ (a straight angle). Why? The classical argument is short and beautiful — it borrows one extra line from outside the triangle and the rest falls out.

The proof in one picture

Take any triangle $ABC$. Through vertex $A$, draw a line parallel to the opposite side $BC$. Call it $\ell$.

ℓ ∥ BC A B C γ α β β γ
The dashed line $\ell$ through $A$ is parallel to $BC$. The angle at $C$ reappears at $A$ on the left; the angle at $B$ reappears on the right.

Now look at vertex $A$. Three angles sit along the line $\ell$:

  • The angle on the left is equal to $\gamma$ (the angle at $C$). Reason: $AC$ is a transversal cutting the parallel lines $\ell$ and $BC$, and these two angles are alternate interior angles.
  • The angle in the middle is $\alpha$ (the original angle of the triangle at $A$).
  • The angle on the right is $\beta$ (the angle at $B$) — same alternate-interior-angle argument, this time with $AB$ as transversal.

Those three angles together fill up the straight line $\ell$, so they sum to $180°$:

$$ \gamma + \alpha + \beta = 180° $$

And that's exactly the claim, with the letters reshuffled. The angle-sum theorem isn't really about triangles — it's about parallel lines wearing a triangle costume.

Euclid only

This proof leans on the existence of a unique parallel line through $A$ — Euclid's parallel postulate. On a curved surface (the surface of a sphere, the inside of a saddle) that postulate fails and triangles don't sum to $180°$. The angles of a triangle on a globe can sum to anything from $180°$ up to $540°$. Section 6 of the pitfalls digs into this.

5. Congruence rules

Two triangles are congruent when one can be slid, rotated, or reflected onto the other so they coincide exactly. Same shape, same size, just possibly facing a different direction.

The remarkable fact is that you don't need to verify all six measurements (three sides, three angles) to confirm congruence. The right three — chosen carefully — are enough. There are four safe patterns:

RuleWhat you checkWhy it works
SSS All three sides equal Three side lengths determine a triangle uniquely (up to reflection).
SAS Two sides and the angle between them equal The angle fixes how the two known sides splay apart; the third side is then forced.
ASA Two angles and the side between them equal The two angles set both directions from the known side; the other two sides meet at a forced point.
AAS Two angles and a non-included side equal Equivalent to ASA: knowing two angles tells you the third (angle-sum), so you're really back to ASA.

Notice what's not on this list: AAA, and SSA.

AAA (all three angles equal) only proves the triangles are similar, not congruent — they have the same shape but possibly different sizes. A small equilateral and a giant equilateral both have three $60°$ angles.

SSA (two sides and a non-included angle) is the famous ambiguous case. The angle and one side determine a ray; the other side then has to reach from the end of the first side to land somewhere on that ray — and depending on its length, it might land at two different spots, one spot, or none. Without more information you can't tell which triangle is meant.

Pitfall

SSA is not a congruence rule. If a problem hands you two sides and an angle that isn't between them, you don't yet have enough to pin the triangle down. You need either the included angle (turning it into SAS) or another side or angle.

6. The triangle inequality

Not every triple of positive numbers makes a triangle. Try to build one with sides $1$, $2$, and $5$: lay the side of length $5$ flat, then try to attach sides of length $1$ and $2$ that meet above it. They can't — they're too short to reach each other.

The rule is:

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

For any three sides of a real triangle, each side must be strictly less than the sum of the other two. Equivalently: the longest side must be strictly less than the sum of the other two.

Why? Picture trying to walk from vertex $B$ to vertex $C$. The direct route is the side of length $a$. The detour through vertex $A$ is the sum $b + c$. The direct route is by definition the shortest path between two points, so

$$ a < b + c $$

and similarly for the other two sides. The inequality is just the statement "a straight line is the shortest distance between two points" turned into algebra.

What happens at the boundary, when $a + b = c$ exactly? The two short sides line up end-to-end and lie flat along the long side. The "triangle" collapses into a single line segment — a degenerate triangle with zero area and three angles of $0°$, $0°$, $180°$. Most definitions exclude this case, which is why the inequality is strict ($>$, not $\geq$).

A quick test

To check whether three lengths can form a triangle, you only need to check one inequality: the one involving the longest side. If the two shorter sides sum to more than the longest, all three inequalities are automatically satisfied.

7. Beyond the basics

A handful of further results come up constantly. None of them is hard once you have the angle-sum theorem and the congruence rules, but each one earns its own name because it gets reused so often.

The exterior angle theorem

At any vertex of a triangle, extend one side past the vertex. The angle that opens up outside the triangle, between the extension and the other side at that vertex, is called an exterior angle. The two interior angles at the other two vertices are called the remote interior angles.

The theorem is short:

An exterior angle equals the sum of the two remote interior angles.

Why? The exterior angle and its adjacent interior angle together form a straight line, so they sum to $180°$. The three interior angles also sum to $180°$. Subtract:

$$ \text{exterior} = 180° - \alpha_{\text{adjacent}} = \beta + \gamma $$

That's the whole proof. As a corollary, an exterior angle is always strictly greater than either of its remote interior angles — useful for bounding arguments.

Similarity

Two triangles are similar ($\triangle ABC \sim \triangle DEF$) when they have the same shape but possibly different sizes. Equivalently: corresponding angles are equal and corresponding sides are in the same ratio:

$$ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} $$

Similarity has its own short list of detection rules, parallel to the congruence rules:

  • AA — two pairs of equal angles. (The third pair is then forced by the angle-sum theorem, so AAA collapses to AA.)
  • SAS similarity — two pairs of sides in the same ratio with the included angle equal.
  • SSS similarity — all three pairs of sides in the same ratio.

Congruence is the special case of similarity where the ratio is $1$. Every fact about congruent triangles has a similarity-scaled cousin.

The Hypotenuse-Leg rule (right triangles only)

Right triangles get one bonus congruence rule, HL (Hypotenuse-Leg): if two right triangles have equal hypotenuses and one pair of equal legs, they are congruent.

Why does this work when generic SSA doesn't? Because the right angle is fixed. The Pythagorean theorem (covered next topic) then forces the third side: if hypotenuse $c$ and leg $a$ match, the other leg $b = \sqrt{c^2 - a^2}$ is determined. SSA's ambiguity vanishes the moment the known angle is $90°$.

The two special right triangles

Two right triangles appear so often — in trigonometry, in mechanical drawings, in any problem involving regular polygons — that their side ratios are worth memorizing.

NameAnglesSide ratio (short : long : hypotenuse)Comes from
$45$-$45$-$90$ $45°$, $45°$, $90°$ $1 : 1 : \sqrt{2}$ Half of a square cut along its diagonal.
$30$-$60$-$90$ $30°$, $60°$, $90°$ $1 : \sqrt{3} : 2$ Half of an equilateral triangle cut along an altitude.

For the $30$-$60$-$90$: the side opposite the $30°$ angle is the shortest (call it $1$), the hypotenuse opposite the $90°$ angle is twice as long ($2$), and the remaining leg opposite the $60°$ angle is $\sqrt{3}$ by Pythagoras.

The four classical triangle centers

Every triangle has four canonical "center" points, each defined by three lines that meet at a single point. The concurrence isn't obvious — that all three meet is the content of each theorem.

CenterBuilt fromWhat it is
Centroid Three medians (vertex to midpoint of opposite side) The triangle's center of mass. Divides each median in a $2{:}1$ ratio, with the longer piece on the vertex side.
Incenter Three angle bisectors Center of the largest circle that fits inside the triangle (the incircle). Equidistant from all three sides.
Circumcenter Three perpendicular bisectors of the sides Center of the unique circle through all three vertices (the circumcircle). Equidistant from all three vertices.
Orthocenter Three altitudes (vertex perpendicular to opposite side) Has no equidistance interpretation, but together with the centroid and circumcenter it lies on a single line — the Euler line.
Where the centers sit

In an acute triangle, all four centers sit inside the triangle. In a right triangle, the circumcenter lands exactly on the midpoint of the hypotenuse and the orthocenter sits at the right-angle vertex. In an obtuse triangle, the circumcenter and orthocenter both move outside the triangle. The centroid and incenter are always interior.

8. Playground: build a triangle

Set three side lengths and watch a triangle assemble itself — or refuse to, when the triangle inequality fails. Use the presets to jump to the classic shapes, then drag the sliders to feel where the boundaries are.

3.0, 4.0, 5.0
valid triangle
3.0
4.0
5.0
Angle A (opp. a) 36.87°
Angle B (opp. b) 53.13°
Angle C (opp. c) 90.00°
By sides Scalene
By angles Right
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Try it

Slide $c$ up toward $a + b$ and watch the triangle flatten — at the boundary $c = a + b$ it collapses into a line segment. Push past, and the warning appears: the two short sides can't reach across the long one. This is the triangle inequality made visible.

9. Common pitfalls

SSA is not a congruence rule

Side-Side-Angle, with the angle not between the two known sides, leaves up to two valid triangles. Don't treat it as a congruence proof. The mnemonic some students learn — that SSA "spells a word you shouldn't say" — is crude but memorable for a reason.

"Isosceles" includes "equilateral"

An equilateral triangle has at least two equal sides — three, in fact — so it satisfies the isosceles definition. Theorems proven for isosceles triangles automatically apply to equilateral ones. Don't write "isosceles but not equilateral" unless you specifically need to exclude the equilateral case.

Angle-sum = $180°$ is Euclidean only

On a sphere (like the Earth's surface), a triangle drawn between three points has angles that sum to more than $180°$. Connect the North Pole to two points on the equator a quarter-turn apart, and you get a triangle with three $90°$ angles — total $270°$. The $180°$ rule depends on the flatness of the plane.

Three sides don't always make a triangle

Before computing area or angles from three side lengths, check the triangle inequality. If two short sides sum to less than (or equal to) the longest, no triangle exists — and any formula that pretends otherwise (Heron's formula, the law of cosines) will return nonsense or a degenerate answer.

10. Worked examples

Try each one yourself before opening the solution. The goal isn't the final number — it's seeing whether your reasoning follows the same shape as the canonical argument.

Example 1 · Find the third angle of a triangle with angles $42°$ and $73°$

By the angle-sum theorem, $\alpha + \beta + \gamma = 180°$. Plug in the two known angles:

$$ 42° + 73° + \gamma = 180° $$

Solve:

$$ \gamma = 180° - 42° - 73° = 65° $$

Since every angle is below $90°$, this is an acute triangle. With three different angles, it's also scalene.

Example 2 · Can sides $4$, $7$, and $12$ form a triangle?

Test the triangle inequality on the longest side ($12$):

$$ 4 + 7 = 11 \;\not>\; 12 $$

The sum of the two shorter sides is less than the longest. No triangle exists with these side lengths. The two short sides aren't long enough to bridge the gap and meet above the long side.

Example 3 · An isosceles triangle has a vertex angle of $40°$. What are the base angles?

The two base angles are equal — call each one $\beta$. The angle-sum theorem gives:

$$ 40° + \beta + \beta = 180° $$ $$ 2\beta = 140° $$ $$ \beta = 70° $$

So the triangle has angles $40°$, $70°$, $70°$. All angles less than $90°$, so this is an acute isosceles triangle.

Example 4 · Are $\triangle ABC$ and $\triangle DEF$ congruent given $AB = DE$, $BC = EF$, and $\angle B = \angle E$?

Look at the configuration: sides $AB$ and $BC$ meet at vertex $B$, and $\angle B$ is the angle between them. Likewise for $\triangle DEF$, $\angle E$ sits between sides $DE$ and $EF$.

This is the SAS pattern: two sides and the angle between them. So yes, $\triangle ABC \cong \triangle DEF$ by SAS.

Watch what would happen if the angle were $\angle A$ instead of $\angle B$ — that's $\angle A$ between $AB$ and $AC$, but we don't know $AC$. With $AB$, $BC$, and $\angle A$ we'd be in SSA territory: ambiguous, not a congruence proof.

Example 5 · Classify the triangle with sides $5$, $5$, $5\sqrt{2}$

By sides. Two sides equal ($5 = 5$), the third different — this is isosceles (not equilateral, since $5 \neq 5\sqrt{2}$).

By angles. The two equal sides face equal base angles; call each $\beta$. The remaining vertex angle, opposite the side of length $5\sqrt{2}$, call $\alpha$. We'll use a fact from the Pythagorean theorem: if $a^2 + b^2 = c^2$, the triangle is right-angled at the vertex opposite $c$. Check:

$$ 5^2 + 5^2 = 25 + 25 = 50 = (5\sqrt{2})^2 $$

So $\alpha = 90°$, and the base angles are then $\beta = \tfrac{180° - 90°}{2} = 45°$.

The triangle is a right isosceles triangle (also called a $45$-$45$-$90$ triangle) — both classifications applying at once.

Sources & further reading

The properties above are standard Euclidean geometry, settled for over two thousand years. The references below are where to turn for proofs we glossed over, harder problems, and the picture-rich version of each idea.

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