Angles in the Same Segment

Same arc, same side, same angle.

Two angles on the same side of a chord

A chord AB cuts a circle into two segments, one on each side of it. Choose two points P and Q on the circle in the same segment, and join each of them to A and to B. The angles APB and AQB both stand on the arc AB, and both look at it from the same side of the chord. They are called angles in the same segment.

The theorem says that angles in the same segment are equal.

xxABPQ

P and Q are both above the chord AB, in the same segment. The angles APB and AQB are equal.

Move one of them

Keep Q where it is and slide P along the upper arc. The angle at P never changes, and it always reads the same as the angle at Q.

Drag P anywhere on the arc above the chord. It reads 70°, the same as Q.

Why: both are half of the same angle

Join A and B to the center O. The angle AOB at the center stands on the arc AB, and so do the angles at P and at Q. By the angle at the center theorem, each angle at the circumference is half of the angle at the center: APB = ½ AOB, and AQB = ½ AOB. Two angles that are each half of the same angle are equal.

In the figures, the angle AOB at the center is 140°, so the angle at every point on the upper arc is 140° ÷ 2 = 70°.

2xxxABPQ

The angle at the center is 2x. The angles at P and at Q each stand on the same arc, so each is half of it, x.

The same segment, not the other one

The two points must be on the same side of the chord. A point R on the short arc, below the chord, stands on the other arc, the long one, so its angle is half of a different angle at the center, and it is not equal to the angles at P and Q.

In the figure below, the angle AOB at the center is 120°, so the angle at every point on the long arc is 60°. Drag the point across the chord onto the short arc and the angle becomes 120°. The two angles, one from each segment, add to 180°: that is the next idea, the angles of a quadrilateral drawn in a circle.

ABP60°

every P on this arc gives the same angle, because ∠APB = ½ × ∠AOB for all of them

Drag P onto the short arc

A and B are fixed, with an angle of 120° between them at the center. On the long arc the angle at P is ½ × 120° = 60° wherever P is. On the short arc it is 180° − 60° = 120°.

Crossed chords

The theorem is often used where two chords cross. Take four points on a circle and join them so that the chords AC and BD cross. The angles ACB and ADB both stand on the arc AB from the same side, so they are equal. The angles CAD and CBD both stand on the arc CD from the same side, so they are equal too.

In the figure, ACB = ADB = 65° and CAD = CBD = 50°. Call the point where the chords cross X. The triangles AXD and BXC then have two pairs of equal angles, 50° and 65°, so their third angles are equal as well, and the two triangles are similar.

65°65°ABCD

The chords AC and BD cross. The angles at C and D stand on the arc AB and are both 65°. The angles marked at A and B stand on the arc CD and are both 50°.

The usual mistakes

Taking 180° − x. If the angle at P is 70°, the angle at Q in the same segment is 70°, not 110°. 180° − x is the angle in the other segment.

Doubling. 2x is the angle at the center. Q is on the circle, like P, so its angle is x.

Pairing angles that stand on different arcs. Check that both angles open toward the same two points, and that their corners are on the same side of the chord joining them.

Worked example: Security Cameras on the Wall of a Round Hall

Question A round hall has center O. Its doorway runs from A to B along the wall, and angle AOB is 84°. A security camera C is fixed to the wall on the far side of the hall from the doorway. A second camera D is fixed to the wall at the point directly opposite A, so that AD is a diameter of the hall. (a) What angle does the doorway fill in the view of camera C, that is, what is angle ACB? (b) Find angle DAB, the angle between the diameter AD and the doorway AB.

  1. 1.Angle AOB at the center and angle ACB at the wall both stand on the arc AB, the doorway. The angle at the center is twice the angle at the circumference, so ACB = 12 × 84° = 42°. (a) The doorway fills 42° of camera C's view.

    84 degOABC: 42 degDdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 deg
    84 degOABC: 42 degDdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 deg
    (a) The angle at the center is twice the angle at the wall: ACB = 12 × 84° = 42°.

  2. 2.Camera D is on the same side of the doorway as camera C, so angles ACB and ADB are angles in the same segment: ADB = 42° as well.

    84 degOABC: 42 degD: 42 degdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 degD on the same arc as C: ADB = 42 deg
    84 degOABC: 42 degD: 42 degdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 degD on the same arc as C: ADB = 42 deg
    D is on the same side of the doorway as C, so angles in the same segment give ADB = 42°.

  3. 3.AD is a diameter, so angle ABD, the angle in a semicircle, is 90°.

    84 degOAB: 90 degC: 42 degD: 42 degdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 degD on the same arc as C: ADB = 42 degAD is a diameter: ABD = 90 deg
    84 degOAB: 90 degC: 42 degD: 42 degdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 degD on the same arc as C: ADB = 42 degAD is a diameter: ABD = 90 deg
    AD is a diameter, so the angle in the semicircle at B is 90°.

  4. 4.(b) In triangle ABD the angles add to 180°: DAB = 180° − 90° − 42° = 48°. Check: triangle OAB is isosceles with OA = OB, so OAB = 12(180° − 84°) = 48°, and OA lies along the diameter AD.

    84 degOA: 48 degB: 90 degC: 42 degD: 42 degdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 degD on the same arc as C: ADB = 42 degAD is a diameter: ABD = 90 degDAB = 180 − 90 − 42 = 48 deg
    84 degOA: 48 degB: 90 degC: 42 degD: 42 degdoorwayangle at the center = 2 × angle at the wallACB = 84/2 = 42 degD on the same arc as C: ADB = 42 degAD is a diameter: ABD = 90 degDAB = 180 − 90 − 42 = 48 deg
    (b) In triangle ABD, DAB = 180° − 90° − 42° = 48°.

Answer: (a) 42°; (b) 48°

Common mistakes

  • Giving 84° as the angle at the camera. The doorway makes 84° at the center of the hall; from the wall, further away, it fills only half of that.
  • Doubling instead of halving, to get 168°. The angle at the center is the larger one, so the angle at the wall is found by halving 84°.

More congruence, similarity and circle theorems problems, worked step by step →

Practice Angles in the Same Segment in the app