What is the sum of the interior angles of a polygon?

For a polygon with n sides the interior angles add up to (n minus 2) times 180 degrees: 180 for a triangle, 360 for a quadrilateral, 540 for a pentagon, 720 for a hexagon.

The diagonals from one vertex cut the polygon into n minus 2 triangles, and each triangle contributes 180 degrees. The triangles' angles fit together to make exactly the polygon's angles.

Does the formula work for non-convex polygons?

Yes, the sum is still (n minus 2) times 180 degrees, counting the reflex angles as larger than 180 degrees. The proof on this page cuts from one vertex, which only works for convex polygons; a non-convex one can be cut into triangles in a different way.

Polygon Angle Sum

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original pentagon. Learn more about the controls
Drag any vertex at any step, keeping the pentagon convex; the diagonals, the three triangles and the readout follow, and the five angles keep adding up to $540°$ . Learn more about why it holds
Labels toggles the point names, Measures writes each angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Draw every diagonal from one vertex of the pentagon: it falls into three triangles. Each triangle has an angle sum of 180°, and the triangles’ angles fit together to make exactly the five angles of the pentagon, so those add up to 3 × 180° = 540°. For any convex polygon with n sides the same cut gives n − 2 triangles: (n − 2) · 180°. Drag any vertex, keeping the shape convex: the sum stays 540°.Polygon angle sum

DisplayLabelsMeasures

Drag any vertex (keep the pentagon convex)

α + β + γ + δ + ε = 540.0° = (5 − 2) · 180°

Step 1 / 5

Step explanations

1A convex pentagon ABCDE

Given
Five sides, five interior angles α, β, γ, δ, ε. To prove: they add up to 540° = (5 − 2) · 180°, and for an n-gon to (n − 2) · 180°.

Convex: every angle below 180°, so the diagonals stay inside. Learn more about the statement

Getting Started with the Visualizer

The Statement

The Diagonals

Three Triangles

The Angles Fit Together

Why It Holds

Dragging the Figure

Consequences

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 6

Step through the proof, then drag the vertices of the pentagon to see that its five angles keep adding up to

540°

whatever its shape. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

1234567

1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each angle's size beside its arc

3The vertices — all five can be dragged, as long as the pentagon stays convex

4Readout — the line under the figure: the sum of the five angles, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original pentagon

6Speed and step counter — how fast Play advances, and which of the 5 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 5 of 5

Proof?Polygon angle sumDisplayLabelsMeasuresDrag any vertex (keep the pentagon convex)

α + β + γ + δ + ε = 540.0° = (5 − 2) · 180°

← Back↻ ReplayNext →ResetNormal ▾Step 5 / 5

Step explanations

1A convex pentagon ABCDE

Given
Five sides, five interior angles α, β, γ, δ, ε. To prove: they add up to 540° = (5 − 2) · 180°, and for an n-gon to (n − 2) · 180°.

2Diagonals from A: three triangles

Construction
The two diagonals AC and AD from the vertex A cut the pentagon into the triangles ABC, ACD and ADE. In a convex polygon every diagonal from one vertex lies inside, so the n − 2 triangles fill it exactly.

33 × 180° = 540°

Triangle angle sum
The angles of each triangle add up to 180°, so the nine angles of the three triangles add up to 3 × 180° = 540°.

4The triangles’ angles make the pentagon’s angles

Angles fit together
At A the three triangle angles sit side by side and make α; at C and at D two triangle angles make γ and δ; at B and E a single triangle angle is the whole of β and ε. Nothing is left over and nothing is counted twice.

5α + β + γ + δ + ε = 540° = (n − 2) · 180°

Generalise
The pentagon’s angles are exactly the triangles’ angles regrouped, so they add up to 540°. With n sides the diagonals from one vertex give n − 2 triangles, hence (n − 2) · 180°. ∎

12345

1The diagonals — AC and AD from the vertex A, inside the convex pentagon

2The three triangles — ABC, ACD, ADE in three colours: 180° each

3The angle at A — three triangle angles side by side make α; two make γ and δ, one makes β and ε

4Justification tags — Given, Construction, Triangle angle sum, Angles fit together, Generalise

5Replay — at the last step Play becomes Replay and Next is off

The Statement

A polygon with

n

sides has

n

vertices and

n

interior angles. The theorem gives their sum:

\text{sum of the interior angles} = (n - 2) \cdot 180°

For a triangle,

n = 3

, this is the

180°

of the triangle angle sum; for a quadrilateral

360°

; for the pentagon in the player

3 \cdot 180° = 540°

; for a hexagon

720°

, and so on,

180°

more for each extra side. The polygon is taken convex here: every interior angle less than

180°

, so that the diagonals from one vertex lie inside it. The formula holds for non-convex polygons too, but the cutting then needs more care.

The readout under the canvas adds the five angles in degrees live while you drag; the proof itself only uses the triangle angle sum and the way angles fit together at a vertex.

Opening scene, frozen

A convex pentagon with its five interior angles α, β, γ, δ, ε.

The Diagonals

The construction is to draw every diagonal from one vertex. From

A

there are two,

AC

and

AD

(the segments to

B

and

E

are sides, not diagonals), and they cut the pentagon into three triangles:

ABC

ACD

and

ADE

, shaded in three colours in the player. Because the pentagon is convex, both diagonals lie inside it and the three triangles fill it exactly, with no gaps and no overlaps.

Count in general: a vertex of an

n

-gon is joined by sides to its two neighbours and by diagonals to the other

n - 3

vertices, and

n - 3

diagonals cut the polygon into

n - 2

triangles. That number,

n - 2

, is where the formula comes from. For the pentagon,

n - 2 = 3

The diagonals, frozen

AC and AD from the vertex A: three triangles ABC, ACD, ADE in three colours.

Three Triangles

Each of the three triangles has an angle sum of

180°

, by the triangle angle sum theorem. So all nine triangle angles together make

3 \cdot 180° = 540°

This counts the angles of the triangles, not yet the angles of the pentagon; the next step shows they are the same. For an

n

-gon the

n - 2

triangles give

(n - 2) \cdot 180°

in the same way.

Three triangles, frozen

Each triangle has 180°: 3 × 180° = 540° for all nine triangle angles.

The Angles Fit Together

Look at each vertex of the pentagon and ask which triangle angles sit there:

• at

A

, three triangle angles sit side by side along the two diagonals and make up the whole of

\alpha

;
• at

C

and at

D

, where one diagonal ends, two triangle angles make up

\gamma

and

\delta

;
• at

B

and at

E

, a single triangle angle is the whole of

\beta

and

\varepsilon

.

Every triangle angle is used exactly once, and every pentagon angle is completely covered: the nine triangle angles are the five pentagon angles cut into pieces. Adding the pieces vertex by vertex, or triangle by triangle, gives the same total. The player highlights the five pentagon angles with the diagonals running through

\alpha

\gamma

and

\delta

The angles fit together, frozen

The five pentagon angles highlighted with the diagonals through α, γ, δ: the triangle angles regrouped.

Why It Holds

The two counts are of the same angles, so

\alpha + \beta + \gamma + \delta + \varepsilon = 3 \cdot 180° = 540°

which is the statement for the pentagon. The end-of-proof mark ∎ in the log closes the argument. The tags name what was used: the convex polygon, the diagonals from one vertex, the triangle angle sum, and the fitting of angles at each vertex; the last step only replaces

5

n

. The general formula

(n - 2) \cdot 180°

follows because the same cut always gives

n - 2

triangles.

There is a second route through the exterior angles: one exterior angle at each vertex always adds up to

360°

, and each interior angle is

180°

minus its exterior angle, so the interior sum is

n \cdot 180° - 360° = (n - 2) \cdot 180°

. The two proofs are the same fact seen from inside and from outside.

The conclusion, frozen

α + β + γ + δ + ε = 540° = (5 − 2) · 180°.

Dragging the Figure

Drag any vertex: the three triangles change shape, their angles change, and the five pentagon angles change with them, but the readout keeps the total at

540.0°

. Pull

B

outward and

\beta

shrinks while its neighbours grow; push a vertex inward and its angle opens up.

A drag that would make the pentagon non-convex, with an interior angle of

180°

or more, or would flatten one of the three triangles, is refused: the proof's cutting into triangles from one vertex needs a convex figure, and the player keeps the last valid one.

After dragging B and D, frozen

A lopsided but still convex pentagon: the angles have changed, the sum is still 540°.

Consequences

• Regular polygons. When all

n

angles are equal, each is

\frac{(n - 2) \cdot 180°}{n}

60°

for the equilateral triangle,

90°

for the square,

108°

for the regular pentagon,

120°

for the regular hexagon.
• Quadrilaterals.

n = 4

gives

360°

: the angles of every quadrilateral, convex or not, add up to a full turn.
• Which regular polygons tile the plane. Copies of a regular polygon fit round a point exactly when

\frac{(n - 2) \cdot 180°}{n}

divides

360°

, which happens only for

n = 3, 4, 6

.
• Number of diagonals. The same count,

n - 3

diagonals from each vertex, gives

\frac{n(n - 3)}{2}

diagonals in all:

5

for a pentagon,

9

for a hexagon.