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Tangent-Chord Angle Theorem

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 BB or CC round the circle, AA for the radius or OO for the centre; the readout under the canvas shows ∠TAB\angle TAB, ∠ADB\angle ADB and ∠ACB\angle ACB. 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?Draw the diameter AD. The tangent is perpendicular to the radius, so ∠TAD = 90°, and the angle in the semicircle makes ∠ABD = 90°. In the right triangle ABD, ∠ADB = 90° − ∠BAD; and at A, ∠TAB = 90° − ∠BAD as well. So ∠TAB = ∠ADB, and ∠ADB = ∠ACB because both stand on the arc AB. Slide B or C round the circle.Tangent-chord angle
Display
Drag B or C round the circle, A for the radius, O for the centre
θθOABCT
∠TAB = 73°, ∠ADB = 73°, ∠ACB = 73°
Step 1 / 6

Step explanations

1A tangent, a chord and an angle in the alternate segment
Given
The line TA touches the circle at A, and AB is a chord from the point of contact. C is any point of the circle on the other side of AB from T. To prove: ∠TAB = ∠ACB, the angle between the tangent and the chord equals the inscribed angle on the chord in the alternate segment.
Drag B or C round the circle. Learn more about the statement

Getting Started with the Visualizer
The Statement
The Diameter AD
The Angle in the Semicircle
The Complement
Inscribed Angles
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 BB or CC round the circle and watch the three angles 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 — B and C slide round the circle, A sets the radius and turns the tangent, O moves the circle; the hollow points D and T follow
4Readout — the line under the figure: the angles at the tangent, at D and 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
12345
1The tangent and the chord — the line through A touching the circle, and the chord AB; θ between them
2The diameter and the right angles — AD through the centre; square marks at A (tangent) and at B (Thales)
3The three equal angles — θ at A, at D and at C; φ is the helper angle between chord and diameter
4Justification tags — Given, Tangent ⟂ radius, Thales' theorem, Angle sum, Inscribed angles, Combining
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A line touches a circle at AA, and a chord ABAB is drawn from the point of contact. The chord splits the circle into two segments; the one on the far side of the chord from the tangent ray ATAT is called the alternate segment. The tangent-chord angle theorem (Euclid III.32, also the alternate segment theorem) says that the angle between the tangent and the chord equals the inscribed angle on the chord in the alternate segment: for any point CC of that segment,

∠TAB=∠ACB\angle TAB = \angle ACB


It is the limiting case of the inscribed angle theorem: slide one end of a chord towards the other and the chord becomes the tangent, while the inscribed angle on it keeps its value. The proof below does not use limits; it uses a diameter, tangent perpendicular to radius, Thales' theorem and inscribed angles.

In the player BB and CC slide round the circle, AA sets the radius and OO moves the circle; the readout shows the angle at the tangent, the helper angle at DD and the angle at CC.
θθOABCT
Opening scene, frozen

The tangent at A, the chord AB and the angle θ at C in the alternate segment.


The Diameter AD

Draw the diameter from the point of contact AA through the centre OO to the far point DD. The tangent is perpendicular to the radius at the point of contact, so

∠TAD=90°\angle TAD = 90°


Call the angle between the chord and the diameter φ=∠BAD\varphi = \angle BAD. The angle we are after is the rest of the right angle: θ=∠TAB=90°−φ\theta = \angle TAB = 90° - \varphi.
θθφOABCDT
The diameter AD, frozen

Tangent perpendicular to radius: ∠TAD = 90°; φ is the angle between the chord and the diameter.


The Angle in the Semicircle

Join BB to DD. The angle at BB stands on the diameter ADAD, so by Thales' theorem it is right:

∠ABD=90°\angle ABD = 90°


ABDABD is a right triangle with the right angle at BB, the hypotenuse ADAD and the angle φ\varphi at AA.
θθφOABCDT
The angle in the semicircle, frozen

ABD shaded: ∠ABD = 90° by Thales.


The Complement

In the right triangle ABDABD the two acute angles add up to 90°90° (angle sum), so the angle at DD is

∠ADB=90°−φ\angle ADB = 90° - \varphi


But ∠TAB\angle TAB is also 90°−φ90° - \varphi, since together with φ\varphi it fills the right angle TADTAD. Two angles equal to the same thing are equal: ∠ADB=∠TAB=θ\angle ADB = \angle TAB = \theta. The angle between the tangent and the chord has been moved onto the circle, at DD.
θθφθOABCDT
The complement, frozen

∠ADB = 90° − φ = ∠TAB, both marked θ.


Inscribed Angles

The angles at DD and at CC both stand on the arc ABAB, and DD and CC lie on the same side of the chord ABAB: both are in the alternate segment, away from TT. By the inscribed angle theorem two inscribed angles on the same arc are equal:

∠ADB=∠ACB\angle ADB = \angle ACB


The helper point DD has done its work: it carries the angle from the tangent to every point of the alternate segment.
θθφθOABCDT
Inscribed angles, frozen

θ at D and θ at C stand on the same arc AB.


Why It Holds

Chain the two equalities:

∠TAB=∠ADB=∠ACB\angle TAB = \angle ADB = \angle ACB


The first because both are the complement of φ\varphi in a right angle (the tangent's right angle at AA, and the right triangle ABDABD), the second because both stand on the arc ABAB. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, tangent perpendicular to radius, Thales, the angle sum, inscribed angles, and combining.

The theorem also holds for the other angle at AA, between the chord and the other half of the tangent: that angle is 180°−θ180° - \theta, and so is the inscribed angle on ABAB in the segment containing TT, by the cyclic quadrilateral page. So the tangent makes with the chord, on either side, the angle in the segment opposite to it.
θθφθOABCDT
The conclusion, frozen

∠TAB = ∠ADB = ∠ACB.


Dragging the Figure

Drag BB round the circle: the chord swings, the angle between the tangent and the chord changes, and the angle at CC follows it; the helper angle at DD stays equal to both. Drag CC anywhere in the alternate segment: the angle at CC does not change at all, which is the inscribed angle theorem at work. Drag AA to change the radius or OO to move the circle; the tangent turns with AA.

The player keeps BB on the same side of the diameter ADAD as TT, so that the angle between the tangent ray ATAT and the chord is acute and the proof reads as written, and keeps CC in the alternate segment, away from AA and BB. A drag that would break one of these, or carry a point out of the frame, is refused.
θθφθOABCDT
After dragging B, frozen

B slid towards D: a wider angle at the tangent, the same angle at C.


Consequences

• Power of a point by similarity. With the tangent-chord angle, the triangles PATPAT and PTBPTB in the secant and tangent figure are similar by AA, which gives PT2=PA⋅PBPT^2 = PA \cdot PB in two lines; that page proves it by Pythagoras instead.
• Constructing a tangent through a given angle. To draw a circle through two points AA, BB touching a given line at AA, the theorem tells which inscribed angle the circle must carry on ABAB.
• The converse. If a line through AA makes with the chord ABAB the angle in the alternate segment, the line is tangent to the circle; it is used to prove that a line touches a circle without finding the centre.
• Equal tangent angles. The two tangents from a point PP make equal angles with the chord of contact, both equal to the inscribed angle on that chord.
• Mixtilinear and tangent circles. When two circles touch at AA, the common tangent at AA gives, through this theorem, equal angles in the two circles; that is the usual way to prove that a point of tangency sends chords to parallel chords.