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Exterior Angle 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. Drag AA, BB, CC or DD round the circle, keeping them in order; the readout under the canvas shows ε\varepsilon, δ\delta and β+δ\beta + \delta. 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?Extend AB beyond B to E. The exterior angle ε = ∠CBE and the interior angle β = ∠ABC make a straight angle, so ε = 180° − β. ABCD is cyclic, so its opposite angles β and δ add up to 180°, and δ = 180° − β too. So ε = δ: the exterior angle equals the interior opposite angle. Slide any vertex round the circle.Cyclic exterior angle
Display
Drag A, B, C or D round the circle (keep them in order)
εδABCDE
ε = ∠CBE = 89°, δ = ∠ADC = 89°, β + δ = 180°
Step 1 / 5

Step explanations

1A cyclic quadrilateral and an exterior angle
Given
ABCD is inscribed in a circle. The side AB is extended beyond B to E, and ε = ∠CBE is the exterior angle at B. δ = ∠ADC is the interior angle at the opposite vertex D. To prove: ε = δ, the exterior angle equals the interior opposite angle.
Drag A or C: ε and δ change together. Learn more about the statement

Getting Started with the Visualizer
The Statement
A Linear Pair
Opposite Angles
Two Supplements
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 AA or CC round the circle and watch the exterior angle at BB and the interior angle at DD in the readout stay equal. 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 slide round the circle in order; the hollow point E follows on AB extended
4Readout — the line under the figure: ε, δ and β + δ, 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 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
12345
1The exterior angle ε — at B, between BC and AB extended to E
2The opposite angle δ — at D, equal to ε
3The interior angle β — at B, the supplement of both ε and δ
4Justification tags — Given, Linear pair, Cyclic quadrilateral, Subtracting, Combining
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Let ABCDABCD be a quadrilateral inscribed in a circle, and extend the side ABAB beyond BB to a point EE. The angle ε=∠CBE\varepsilon = \angle CBE is the exterior angle at BB, and δ=∠ADC\delta = \angle ADC is the interior angle at DD, the vertex opposite BB. The theorem says

ε=δ\varepsilon = \delta


the exterior angle of a cyclic quadrilateral equals the interior opposite angle. It is the circle's version of the exterior angle theorem for triangles, and it is a restatement of the cyclic quadrilateral theorem, that opposite angles add up to 180°180°, in a form that is often quicker to use.

In the player the four vertices slide round a fixed circle; the readout shows ε\varepsilon, δ\delta and the sum of the opposite angles β+δ\beta + \delta.
εδABCDE
Opening scene, frozen

The cyclic quadrilateral with AB extended to E; ε at B and δ at D marked.


A Linear Pair

At BB the interior angle β=∠ABC\beta = \angle ABC and the exterior angle ε=∠CBE\varepsilon = \angle CBE lie side by side along the straight line through AA, BB and EE. Together they fill a straight angle:

β+ε=180°\beta + \varepsilon = 180°
εδβABCDE
A linear pair, frozen

β and ε along the line A, B, E: β + ε = 180°.


Opposite Angles

The vertices BB and DD are opposite in the quadrilateral. By the cyclic quadrilateral theorem the opposite angles of a quadrilateral inscribed in a circle add up to 180°180°:

β+δ=180°\beta + \delta = 180°


(That page proves it with four isosceles triangles on the radii; it also follows from the inscribed angle theorem, since β\beta and δ\delta stand on the two arcs ADCADC and ABCABC, which together make the whole circle.)
εδβABCDE
Opposite angles, frozen

β and δ, opposite angles of a cyclic quadrilateral: β + δ = 180°.


Two Supplements

Both ε\varepsilon and δ\delta make 180°180° with the same angle β\beta:

ε=180°−β=δ\varepsilon = 180° - \beta = \delta


so the exterior angle at BB equals the interior angle at DD.
εδβABCDE
Two supplements, frozen

ε = 180° − β = δ.


Why It Holds

The exterior angle at a vertex is the supplement of the interior angle there, and in a cyclic quadrilateral the interior opposite angle is the supplement of that same interior angle. Two supplements of one angle are equal:

ε=δ\varepsilon = \delta


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, the linear pair, the cyclic quadrilateral, subtracting and combining.

The converse is a useful test: if an exterior angle of a quadrilateral equals the interior opposite angle, then the opposite angles add up to 180°180°, and by the converse of the cyclic quadrilateral theorem the four vertices lie on one circle.
εδβABCDE
The conclusion, frozen

The exterior angle at B equals the interior angle at D.


Dragging the Figure

Drag any vertex round the circle: the quadrilateral changes shape, the extension at BB turns with AA and BB, and the readout shows ε\varepsilon and δ\delta changing together while β+δ\beta + \delta stays at 180°180°. Both angles depend only on AA and CC: moving DD along its arc, or BB along its arc, changes neither, since every point of an arc sees the chord ACAC under the same angle.

The player keeps the vertices in order round the circle, the sides long enough for the angle marks and the extension inside the frame; a drag that would break one of these is refused.
εδβABCDE
After dragging C, frozen

C slid round the circle: ε and δ changed together, still equal.


Consequences

• Testing for concyclic points. To show that four points lie on a circle it is enough to find one exterior angle equal to the interior opposite angle; this is how many olympiad configurations are proved cyclic.
• Isosceles trapezoids. In a cyclic trapezoid the exterior angle at one end of a base equals the angle at the other end of the same base, which is why every cyclic trapezoid is an isosceles trapezoid.
• Similar triangles from secants. For two secants from a point PP meeting the circle at A,BA, B and D,CD, C, the exterior angle theorem makes the triangles PBCPBC and PDAPDA similar, which gives the two secants theorem PA⋅PB=PC⋅PDPA \cdot PB = PC \cdot PD by similarity.
• The tangent case. When CC slides into BB the side BCBC becomes a tangent and the theorem becomes the tangent-chord angle theorem.
• Triangles. For a triangle the exterior angle equals the sum of the two remote interior angles; for a cyclic quadrilateral it equals the single opposite angle. Both say that the exterior angle is what the rest of the figure forces it to be.