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Angles Around a Point

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag AA, BB, CC or DD round OO, or OO itself; the readout under the canvas shows the four angles and their sum. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Four rays from O make the angles α, β, γ, δ all the way round. Extend AO beyond O to A′: the line AA′ splits the full turn into two straight angles. On one side α + β + γ₁ = 180°, on the other γ₂ + δ = 180°, and γ₁ + γ₂ = γ, so α + β + γ + δ = 360°. Drag any of A, B, C, D round O.Angles around a point
Display
Drag A, B, C or D round O (keep them in order), or O itself
αβγδOABCD
α = 70°, β = 69°, γ = 101°, δ = 120°, sum = 360°
Step 1 / 6

Step explanations

1Four angles around a point
Given
Four rays OA, OB, OC, OD go out from the point O, in that order round it, and cut the plane around O into the angles α = ∠AOB, β = ∠BOC, γ = ∠COD and δ = ∠DOA. To prove: α + β + γ + δ = 360°, the angles around a point make a full turn.
Drag a ray: two angles change, the sum does not. Learn more about the statement

Getting Started with the Visualizer
The Statement
Extending One Ray
One Side of the Line
The Other Side
Adding the Two Sides
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag any of the four rays round OO and watch the four angles change while their sum in the readout stays at 360°360°. 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 marked angle's size beside its arc
3Draggable points — A, B, C and D move freely round O (in order), O moves the whole figure; the hollow point A′ follows on the extension of AO
4Readout — the line under the figure: the four angles and their sum, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 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 6 of 6
12345
1The four angles — α, β, γ, δ round O, each between two neighbouring rays
2The extension — the dashed ray OA′ continuing AO, which cuts γ into γ₁ and γ₂
3The two pieces of γ — γ₁ and γ₂ on either side of the dashed line, inside the larger arc of γ
4Justification tags — Given, Construction, Straight angle (twice), Adding, Substituting
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Several rays from one point OO cut the plane around OO into angles, one between each pair of neighbouring rays. The theorem says that these angles make a full turn:

α+β+γ+δ=360°\alpha + \beta + \gamma + \delta = 360°


for the four rays in the player, and the same for any number of rays. It is the companion of the straight angle: two angles on a line make 180°180°, the angles all the way round make twice that. Together with the vertical angles and the bisectors of a linear pair pages it covers the basic angle facts at a point.

The proof reduces the full turn to two straight angles by one construction. In the player AA, BB, CC, DD and OO are free; the readout shows the four angles and their sum.
αβγδOABCD
Opening scene, frozen

Four rays from O and the four angles α, β, γ, δ between them.


Extending One Ray

Pick one ray, OAOA, and continue it beyond OO to a point A′A' (dashed). The three points AA, OO, A′A' lie on one straight line, and that line passes through one of the four angles, γ\gamma, cutting it into two parts:

γ1=∠COA′,γ2=∠A′OD,γ1+γ2=γ\gamma_1 = \angle COA', \qquad \gamma_2 = \angle A'OD, \qquad \gamma_1 + \gamma_2 = \gamma


Every other angle lies wholly on one side of the line.
αβγδγ₁γ₂OABCDA′
Extending one ray, frozen

AO continued to A′ (dashed), cutting γ into γ₁ and γ₂.


One Side of the Line

Look at the side of the line AA′AA' that contains BB and CC. Going round from the ray OAOA to the ray OA′OA' one meets α\alpha, β\beta and γ1\gamma_1 in turn, side by side; together they fill the straight angle AOA′AOA':

α+β+γ1=180°\alpha + \beta + \gamma_1 = 180°
αβγδγ₁γ₂OABCDA′
One side of the line, frozen

α, β and γ₁ lit: they fill the straight angle AOA′.


The Other Side

On the other side of the line, going round from OA′OA' back to OAOA, one meets γ2\gamma_2 and δ\delta. They fill the other straight angle:

γ2+δ=180°\gamma_2 + \delta = 180°


Two straight angles, one on each side of the line, and every piece of the full turn is in exactly one of them.
αβγδγ₁γ₂OABCDA′
The other side, frozen

γ₂ and δ lit: the other straight angle.


Adding the Two Sides

Add the two equations:

α+β+γ1+γ2+δ=180°+180°=360°\alpha + \beta + \gamma_1 + \gamma_2 + \delta = 180° + 180° = 360°


and put the two pieces of γ\gamma back together, γ1+γ2=γ\gamma_1 + \gamma_2 = \gamma.
αβγδγ₁γ₂OABCDA′
Adding the two sides, frozen

180° + 180°, with γ₁ + γ₂ = γ.


Why It Holds

The result is

α+β+γ+δ=360°\alpha + \beta + \gamma + \delta = 360°


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, two straight angles, an addition and a substitution. Nothing in the proof used the number of rays: with any rays from OO, one ray extended to a line gives two straight angles, and the angles on each side fill one of them.

The one thing to check is that the extension OA′OA' really falls inside one of the angles and not along another ray; if it lands on a ray, the two straight angles are already made of whole angles and there is nothing to split.
αβγδγ₁γ₂OABCDA′
The conclusion, frozen

α + β + γ + δ = 360°.


Dragging the Figure

Drag AA, BB, CC or DD round OO: two neighbouring angles change, one growing as the other shrinks, and the sum in the readout stays at 360°360°. Drag AA and the dashed extension turns with it, cutting γ\gamma at a different place; drag OO to move the whole figure.

The player keeps the four rays in their order round OO, every angle at least 18°18°, and the extension of OAOA inside the angle γ\gamma so that the proof reads as written. A drag that would swap two rays, close an angle or carry a point out of the frame is refused.
αβγδγ₁γ₂OABCDA′
After dragging B, frozen

B swung down: α larger, β smaller, the sum still 360°.


Consequences

• Exterior angles of a polygon. Walking once round a convex polygon turns through every exterior angle in turn, and a full turn is 360°360°: that is the picture behind the polygon exterior angle sum.
• Regular divisions. nn equal angles round a point are 360°/n360°/n each: 90°90° for four, 60°60° for six, 45°45° for eight; this is how the regular polygons and the tilings by equilateral triangles, squares and hexagons are set out round a vertex.
• Vertical angles. Two crossing lines make four angles round their crossing point; the straight angles on each line show that opposite ones are equal (vertical angles).
• The angle sum of a polygon by fanning. Joining an interior point to every vertex of a polygon makes nn triangles whose angles at that point add up to 360°360°; subtracting it from n⋅180°n \cdot 180° gives the polygon angle sum (n−2)⋅180°(n - 2) \cdot 180° a second way.
• Clock angles. The hour marks divide the full turn into twelve angles of 30°30°; every clock-angle puzzle rests on this theorem.