▶ 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
Slide B or C round the circle, drag A to change the radius or O to move the circle; the readout under the canvas keeps the angle at O at twice the angle at C. 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 the diameter from C through the centre O. It splits the angle at C into α and β, and the angle at O into two parts. Triangle OCA is isosceles, so ∠OAC = α, and the exterior angle at O is α + α = 2α; the same on the other side gives 2β. So the angle at the centre is 2(α + β), twice the angle at the circle. Slide A, B or C round the circle.Inscribed angle theorem
Display
Drag B or C round the circle, A for the radius, O for the centre
∠AOB 126.2° = 2 × ∠ACB 63.1°
Step 1 / 6
Step explanations
1Angle at the circle, angle at the centre
Given A, B, C lie on a circle with centre O. The inscribed angle ∠ACB and the central angle ∠AOB stand on the same arc AB. To prove: ∠AOB = 2 ∠ACB.
Step through the proof, then slide the points round the circle to see that the angle at the centre stays exactly twice the angle at the circle. 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
Proof?Inscribed angle theoremDisplayLabelsMeasuresDrag B or C round the circle, A for the radius, O for the centre
∠AOB 126.2° = 2 × ∠ACB 63.1°
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1Angle at the circle, angle at the centre
Given A, B, C lie on a circle with centre O. The inscribed angle ∠ACB and the central angle ∠AOB stand on the same arc AB. To prove: ∠AOB = 2 ∠ACB.
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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
3Draggable points — B and C slide round the circle, A sets the radius, O moves the circle; the hollow point D is the far end of the diameter through C
4Readout — the line under the figure: the angle at O against twice the angle at C, 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
Proof?Inscribed angle theoremDisplayLabelsMeasuresDrag B or C round the circle, A for the radius, O for the centre
∠AOB 126.2° = 2 × ∠ACB 63.1°
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1Angle at the circle, angle at the centre
Given A, B, C lie on a circle with centre O. The inscribed angle ∠ACB and the central angle ∠AOB stand on the same arc AB. To prove: ∠AOB = 2 ∠ACB.
2The diameter CD through O
Construction Join C to O and extend to D on the circle. OA, OB and OC are radii, equal (one tick each). The diameter splits the angle at C into α + β and the angle at O into ∠AOD + ∠DOB.
3∠OAC = ∠OCA = α
Isosceles base angles OA = OC, so triangle OAC is isosceles with apex O, and its base angles at A and at C are equal.
4∠AOD = α + α = 2α
Exterior angle theorem C, O, D lie on one line, so ∠AOD is an exterior angle of triangle OAC. It equals the sum of the two remote interior angles, α at A and α at C.
5∠DOB = 2β
Same argument On the other side: OB = OC makes triangle OBC isosceles, ∠OBC = ∠OCB = β, and the exterior angle ∠DOB of that triangle equals β + β.
6∠AOB = 2α + 2β = 2 ∠ACB
Adding the parts The angle at the centre is ∠AOD + ∠DOB = 2α + 2β = 2(α + β), and α + β is the angle at the circle. ∎
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1The central angle — at O, in two parts 2α and 2β on either side of the diameter
2The inscribed angle — at C, in two parts α and β; each part at O is twice the matching part at C
3The diameter CD — through the centre to the hollow point D; the line that splits both angles
4Justification tags — Given, Construction, Isosceles base angles, Exterior angle theorem, Same argument, Adding the parts
5Replay — at the last step Play becomes Replay and Next is off
The Statement
Take three points A, B, C on a circle with centre O. The angle ∠ACB, with its vertex on the circle, is an inscribed angle; the angle ∠AOB, with its vertex at the centre, is a central angle. Both stand on the same arc AB, the one that does not contain C. The theorem, Proposition 20 of Book III of Euclid's Elements, says
∠AOB=2⋅∠ACB
the angle at the centre is twice the angle at the circle. Two things follow at once and are worth saying now: the inscribed angle does not depend on where C is on its arc, since ∠AOB does not move; and when AB is a diameter the central angle is 180°, so the inscribed angle is 90°, which is Thales' theorem.
The proof on this page covers the case where the centre lies inside the inscribed angle, the one the player keeps you in. The other cases, centre on a side or outside the angle, use the same two ideas with a subtraction instead of an addition; see Why It Holds. The readout under the canvas shows both angles in degrees, live while you drag.
Opening scene, frozen
A, B, C on the circle; the inscribed angle at C and the central angle at O, each shown in two parts.
The Diameter
The construction is the diameter through C: join C to the centre O and extend the line to the point D on the far side of the circle. In the player D is hollow, because it is built from C and O and follows them.
The diameter does two things at once. It splits the inscribed angle at C into two parts, α=∠ACD on the side of A and β=∠DCB on the side of B; and it splits the central angle at O into ∠AOD and ∠DOB. The proof will show that each part at O is twice the matching part at C. It also brings in the radii OA, OB and OC, all equal because they are radii (one tick each), which is what makes the two triangles OAC and OBC isosceles.
The diameter, frozen
CD through O; the radii OA, OB, OC marked equal.
An Isosceles Triangle
In triangle OAC the sides OA and OC are radii, so they are equal and the triangle is isosceles with apex O. By the isosceles base angles theorem its base angles are equal:
∠OAC=∠OCA=α
The angle at C here is the part α of the inscribed angle on the side of A; the angle at A is new and is marked in the same colour. This is the same opening move as in Thales' theorem, where the two isosceles triangles on the radii did all the work.
The isosceles triangle, frozen
OAC with OA = OC: the base angles at A and at C are both α.
The Exterior Angle
Since C, O and D lie on one line, the angle ∠AOD at O lies outside triangle OAC, between the side OA and the extension OD of the side CO: it is an exterior angle of that triangle. By the exterior angle theorem it equals the sum of the two remote interior angles, which are α at A and α at C:
∠AOD=α+α=2α
This is the whole mechanism of the proof: the isosceles triangle doubles the angle α by producing a second copy of it at A, and the exterior angle collects both copies at the centre. The player highlights the three angles involved.
The exterior angle, frozen
∠AOD is an exterior angle of triangle OAC: α + α = 2α.
The Other Side
The same two steps on the side of B. Triangle OBC has OB=OC, so it is isosceles and ∠OBC=∠OCB=β; and ∠DOB is an exterior angle of triangle OBC at O, so
∠DOB=β+β=2β
Both parts of the central angle are now known: ∠AOD=2α and ∠DOB=2β, each twice the part of the inscribed angle on its own side of the diameter.
The other side, frozen
The same on the side of B: ∠OBC = β and ∠DOB = 2β.
Why It Holds
Add the two parts at O and the two parts at C:
∠AOB=∠AOD+∠DOB=2α+2β=2(α+β)=2⋅∠ACB
which is the statement. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given circle, the diameter through C with its equal radii, the isosceles base angles, the exterior angle theorem, the same two once more, and an addition.
The other two cases need one change. If O lies on a side of the inscribed angle, say on CB, then β=0 and the first half of the proof is already the whole proof: ∠AOB=2α. If O lies outside the inscribed angle, both A and B are on the same side of the diameter CD; the two exterior angles are still 2α and 2β, but now the central angle is their difference, 2β−2α=2(β−α), and the inscribed angle is β−α. Twice, in every case.
The conclusion, frozen
The two parts at O and the two parts at C: ∠AOB = 2α + 2β = 2∠ACB.
Dragging the Figure
Slide C along its arc: α and β trade places, the diameter swings round, and the readout keeps ∠AOB at exactly twice ∠ACB; and since ∠AOB does not change while C moves on its arc, neither does ∠ACB. Slide B to change the arc, or drag A to change the radius and O to move the circle.
The player keeps the figure in the case the proof covers: a drag that would carry C past the diameter through A or B, putting the centre outside the inscribed angle, is refused, as is a drag that would flatten one of the triangles. The angle sum α+β at C is always less than 90° in this case, since 2(α+β) must fit inside a straight angle at O.
After dragging, frozen
C slid along its arc and B moved: α and β have changed, the angle at the centre is still twice the angle at the circle.
Consequences
• Thales' theorem. When AB is a diameter the central angle is 180°, so every inscribed angle on it is 90°; see its own page. • Angles in the same segment. All inscribed angles standing on the same arc are equal, since each is half of the same central angle. This is Euclid III.21 and the key to most circle problems. • Cyclic quadrilaterals. For a quadrilateral inscribed in a circle, two opposite angles stand on arcs that together make the whole circle, so their central angles add up to 360° and the inscribed angles add up to 180°. • Tangent and chord. The angle between a tangent and a chord equals the inscribed angle on the other side of the chord, the limiting case as C slides into the point of tangency.