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Opposite Angles of a Cyclic Quadrilateral

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. Slide BB, CC or DD round the circle, drag AA to change the radius or OO to move the circle; the readout under the canvas keeps both sums of opposite angles at 180°180°. Learn more about why it holds
  3. 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?Join the centre O to A, B, C, D. The four radii are equal, so the four triangles OAB, OBC, OCD, ODA are isosceles, with base angles α, β, γ, δ. Each angle of ABCD is made of two of them, so all four add up to 2(α + β + γ + δ); and the angles of a quadrilateral add up to 360°, so α + β + γ + δ = 180°. That sum is ∠A + ∠C, and also ∠B + ∠D. Slide a vertex round the circle.Cyclic quadrilateral
Display
Drag B, C or D round the circle, A for the radius, O for the centre
OABCD
∠A + ∠C = 180.0°, ∠B + ∠D = 180.0°
Step 1 / 6

Step explanations

1A quadrilateral inscribed in a circle
Given
A, B, C, D lie on a circle with centre O, in that order round it. To prove: ∠A + ∠C = 180°, and likewise ∠B + ∠D = 180°.
The readout shows both sums of opposite angles. Learn more about the statement

Getting Started with the Visualizer
The Statement
The Four Radii
Four Isosceles Triangles
The Angles of the Quadrilateral
The Angle Sum
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 slide the vertices round the circle and watch each pair of opposite angles keep its sum of 180°180°. 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
3Draggable points — B, C and D slide round the circle, A sets the radius, O moves the circle
4Readout — the line under the figure: the two sums of opposite 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 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 angles at A and C — lit at the last step: δ + α and β + γ, all four letters once
2The four radii — equal, one tick each; the apex of every isosceles triangle is O
3The other pair — α + β at B and γ + δ at D, the same four letters again
4Justification tags — Given, Construction, Isosceles base angles, Adding the parts, Polygon angle sum, Adding the parts
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A quadrilateral whose four vertices lie on one circle is called cyclic, and the circle is its circumcircle. Not every quadrilateral is cyclic: a rectangle is, a general parallelogram is not. The theorem, Proposition 22 of Book III of Euclid's Elements, says that in a cyclic quadrilateral ABCDABCD the opposite angles are supplementary:

∠A+∠C=180°,∠B+∠D=180°\angle A + \angle C = 180°, \qquad \angle B + \angle D = 180°


The proof on this page joins the centre to the four vertices and uses two earlier results, the isosceles base angles theorem and the angle sum of a polygon. It covers the case where the centre lies inside the quadrilateral, the one the player keeps you in; a second proof, through the inscribed angle theorem, covers every case at once and is sketched under Why It Holds. The readout under the canvas shows both sums in degrees, live while you drag.
OABCD
Opening scene, frozen

ABCD with its four vertices on the circle; the centre O inside.


The Four Radii

The construction is to join the centre OO to each vertex. The four segments OAOA, OBOB, OCOC, ODOD are radii of the same circle, so they are all equal; the player marks each with one tick. They cut the quadrilateral into four triangles, OABOAB, OBCOBC, OCDOCD, ODAODA, which fit together round OO like slices.

Each of these triangles has two sides that are radii, so each is isosceles with its apex at OO. That is the whole point of the construction: it turns a statement about four angles into a statement about four isosceles triangles, whose base angles are easy to track.
OABCD
The four radii, frozen

OA, OB, OC, OD, all equal: four isosceles triangles round O.


Four Isosceles Triangles

In the isosceles triangle OABOAB the base angles, at AA and at BB, are equal; call them α\alpha. In OBCOBC the base angles at BB and CC are equal; call them β\beta. In OCDOCD they are γ\gamma, at CC and DD; in ODAODA they are δ\delta, at DD and AA. The player draws each pair in its own colour.

Eight marked angles, four letters: every letter sits at two vertices, the two ends of one side of the quadrilateral. Nothing has been assumed about the sizes of α\alpha, β\beta, γ\gamma, δ\delta; they change as you drag, and the argument never needs them.
ααββγγδδOABCD
The base angles, frozen

α at A and B, β at B and C, γ at C and D, δ at D and A.


The Angles of the Quadrilateral

The radius through each vertex splits the angle of the quadrilateral there into two of the marked angles. At AA, the radius OAOA lies between ABAB and ADAD, so

∠A=δ+α,∠B=α+β,∠C=β+γ,∠D=γ+δ\angle A = \delta + \alpha, \quad \angle B = \alpha + \beta, \quad \angle C = \beta + \gamma, \quad \angle D = \gamma + \delta


This is where the case matters. The radius lies between the two sides at every vertex exactly when the centre is inside the quadrilateral; if OO were outside, one of these sums would be a difference instead. Notice the pattern: each letter appears in two neighbouring angles, and opposite angles share no letter.
ααββγγδδOABCD
The angles of ABCD, frozen

Each angle of the quadrilateral is two of the letters; the angles at A and at C are lit.


The Angle Sum

A quadrilateral can be cut into two triangles by a diagonal, so its angles add up to 2⋅180°=360°2 \cdot 180° = 360°; this is the polygon angle sum with n=4n = 4. Adding the four expressions from the previous step,

∠A+∠B+∠C+∠D=2(α+β+γ+δ)=360°\angle A + \angle B + \angle C + \angle D = 2(\alpha + \beta + \gamma + \delta) = 360°


so α+β+γ+δ=180°\alpha + \beta + \gamma + \delta = 180°. The four base angles, one from each isosceles triangle, together make a straight angle.
ααββγγδδOABCD
The angle sum, frozen

All eight marks together: 2(α + β + γ + δ) = 360°.


Why It Holds

Now collect opposite angles: ∠A+∠C=(δ+α)+(β+γ)\angle A + \angle C = (\delta + \alpha) + (\beta + \gamma), which is all four letters once, so it equals 180°180°. In the same way ∠B+∠D=(α+β)+(γ+δ)=180°\angle B + \angle D = (\alpha + \beta) + (\gamma + \delta) = 180°. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given circle, the construction with equal radii, the isosceles theorem, an addition, the polygon angle sum, and an addition once more.

The other proof, which needs no case split, uses the inscribed angle theorem. The angle at AA stands on the arc BCDBCD and the angle at CC stands on the arc DABDAB; each is half the central angle on its arc, and the two central angles together make the full turn, 360°360°. So ∠A+∠C=12⋅360°=180°\angle A + \angle C = \tfrac{1}{2} \cdot 360° = 180°. One of the two central angles may be larger than a straight angle, which the player does not draw; that is why this page shows the radii proof instead.
ααββγγδδOABCD
The conclusion, frozen

∠A + ∠C = δ + α + β + γ = 180°.


Dragging the Figure

Slide any of BB, CC, DD round the circle: the four isosceles triangles change shape, the eight marked angles change size, and the readout keeps both sums at 180°180°. Slide CC towards BB and watch β\beta shrink while γ\gamma grows; the angle at CC changes, and the angle at AA changes by the same amount the other way. Drag AA to change the radius, OO to move the circle.

The player keeps the figure in the case the proof covers: a drag that would carry a vertex past its neighbour, or push the centre outside the quadrilateral, is refused. In that refused case the theorem is still true, as the second proof shows, but one of the four sums in The Angles of the Quadrilateral would become a difference.
ααββγγδδOABCD
After dragging C and D, frozen

C slid towards B: β is small and γ large, the sums are still 180°.


Consequences

• The converse. If a quadrilateral has a pair of opposite angles adding up to 180°180°, it is cyclic: a circle passes through all four vertices. This is the usual way to prove that four points lie on a circle.
• The exterior angle. Extend a side of a cyclic quadrilateral; the exterior angle at that vertex equals the interior angle at the opposite vertex, since both are supplementary to the same angle.
• Rectangles and isosceles trapezoids are cyclic; a parallelogram is cyclic only if it is a rectangle, and a rhombus only if it is a square.
• Ptolemy's theorem. In a cyclic quadrilateral the product of the diagonals equals the sum of the products of opposite sides, AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB \cdot CD + AD \cdot BC.
• Thales again. A rectangle inscribed in a circle has its diagonals as diameters, so each of its right angles stands on a diameter, as Thales' theorem says.