Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools

Angles in the Same Segment

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 CC or DD along the arc, or AA or BB round the circle; the readout under the canvas shows both angles and half the central angle. 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?The chord AB cuts the circle into two segments, and C and D lie in the same one. By the inscribed angle theorem each of the angles ACB and ADB is half the central angle AOB on the same arc. Two halves of one angle are equal, so ∠ACB = ∠ADB. Slide C or D along its arc, or move A and B.Angles in the same segment
Display
Drag C or D along the arc, or A or B round the circle
θ₁θ₂ABCD
θ₁ = ∠ACB = 47°, θ₂ = ∠ADB = 47°, ω/2 = 47°
Step 1 / 5

Step explanations

1Two angles on the same chord
Given
AB is a chord of the circle, and C and D are two points on the same arc AB, the same side of the chord. θ₁ = ∠ACB and θ₂ = ∠ADB are the angles under which C and D see the chord. To prove: θ₁ = θ₂, angles in the same segment are equal.
Drag C or D along the arc: the angle stays. Learn more about the statement

Getting Started with the Visualizer
The Statement
The Central Angle
The First Inscribed Angle
The Second Inscribed Angle
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 CC or DD along the arc, or AA or BB round the circle, and watch the two angles in the readout stay equal to each other and to half the central angle. 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 — C and D slide along the arc, A and B round the circle; the centre O stays fixed
4Readout — the line under the figure: θ₁, θ₂ and half the central angle, 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 two inscribed angles — θ₁ at C and θ₂ at D, both on the chord AB
2The central angle — ω at O between the radii OA and OB
3The chord AB — the common base of the two shaded triangles
4Justification tags — Given, Construction, Inscribed angle theorem (twice), Combining
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A chord ABAB cuts a circle into two segments. Take two points CC and DD on the same arc, in the same segment. The theorem (Euclid III.21) says that they see the chord under the same angle:

∠ACB=∠ADB\angle ACB = \angle ADB


All points of one arc see the chord ABAB under one fixed angle. It is a corollary of the inscribed angle theorem, and it is the form of that theorem used most often: in the intersecting chords and two secants arguments by similarity, in Ptolemy's theorem and in the Simson line.

In the player the four points slide round the circle, with CC and DD kept on the arc on the same side of ABAB as the centre OO; the readout shows both angles and half the central angle.
θ₁θ₂ABCD
Opening scene, frozen

The chord AB and two points C, D on the same arc, with the angles θ₁ and θ₂.


The Central Angle

Join the centre OO to AA and to BB. The angle ω=∠AOB\omega = \angle AOB is the central angle standing on the arc ABAB, the arc on the far side of the chord from CC and DD. Both inscribed angles will be compared with it.
θ₁θ₂ωOABCD
The central angle, frozen

OA and OB drawn: the central angle ω on the arc AB.


The First Inscribed Angle

The angle ACBACB at the circumference and the angle AOBAOB at the centre stand on the same arc ABAB. The inscribed angle theorem says that the angle at the circumference is half the angle at the centre:

θ1=∠ACB=ω2\theta_1 = \angle ACB = \frac{\omega}{2}
θ₁θ₂ωOABCD
The first inscribed angle, frozen

ACB shaded: θ₁ = ω/2.


The Second Inscribed Angle

The point DD is on the same arc as CC, so the angle ADBADB stands on the same arc ABAB and the same theorem applies:

θ2=∠ADB=ω2\theta_2 = \angle ADB = \frac{\omega}{2}
θ₁θ₂ωOABCD
The second inscribed angle, frozen

ADB shaded: θ₂ = ω/2.


Why It Holds

Both angles are half of the same central angle:

θ1=ω2=θ2\theta_1 = \frac{\omega}{2} = \theta_2


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, the inscribed angle theorem twice, and combining. The player shows the case with the centre inside both angles; when it is outside, the inscribed angle theorem still gives half the central angle (by subtraction instead of addition), so the result is the same.

Points on the other arc, in the other segment, see the chord under the supplementary angle 180°−θ1180° - \theta_1: with CC on one arc and EE on the other, ACBEACBE is a cyclic quadrilateral and its opposite angles add up to 180°180°.
θ₁θ₂ωOABCD
The conclusion, frozen

θ₁ = θ₂: the same angle from every point of the arc.


Dragging the Figure

Drag CC or DD along the arc: the lines from it to AA and BB swing, but the angle between them, shown in the readout, does not change at all. Drag AA or BB to change the chord: both angles change, together, and stay equal to half the central angle.

The player keeps CC and DD on the arc on the centre's side of the chord, with the centre inside both angles, and keeps the points apart and in the frame; a drag that would break one of these is refused.
θ₁θ₂ωOABCD
After dragging D, frozen

D slid along the arc: a different point, the same angle.


Consequences

• The converse: a locus. The points on one side of ABAB that see it under a fixed angle θ\theta fill an arc of a circle through AA and BB. For θ=90°\theta = 90° this is the semicircle of Thales' theorem.
• Concyclic points. If CC and DD on the same side of ABAB see it under equal angles, then AA, BB, CC, DD lie on one circle; this is the test behind many cyclic-quadrilateral proofs, alongside the exterior angle test.
• Similar triangles from chords. Two chords ACAC and BDBD crossing at PP make the triangles PABPAB and PDCPDC similar, since ∠PAB=∠PDC\angle PAB = \angle PDC (same segment); that gives PA⋅PC=PB⋅PDPA \cdot PC = PB \cdot PD, the intersecting chords theorem.
• The tangent case. When DD slides into BB, the line DBDB becomes the tangent at BB and the equal angle becomes the tangent-chord angle.
• Drawing a circle through three points with a given angle. To find all points that see a segment under 60°60°, draw the arcs through its ends on which the inscribed angle is 60°60°; they are used to locate a ship from the angle between two lighthouses.