Year 11 Extended · Unit 1 Right-angled triangles and circle geometry · Lesson 5 of 7

The angle at the centre and its consequences: one proof, every case

A single result about the angle at the centre gives the angle in a semicircle, angles in the same segment, the cyclic quadrilateral and the alternate segment. This lesson proves it whether the centre is inside the angle, on its side or outside it, and then lets the consequences follow.

Geometry and TrigonometryNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: the base angles of an isosceles triangle with apex angle 50°.
    Answer65° each, because (180 − 50) ÷ 2.
  2. From Year 9: the exterior angle of a triangle whose two opposite interior angles are 28° and 35°.
    Answer63°, the sum of the two opposite angles.
  3. From Year 9: the angles of a quadrilateral. What do they add to?
    Answer360°.
Words used in this lesson
major arc
the longer of the two arcs between two points on a circle. For example: the long way round.
minor arc
the shorter of the two arcs between two points on a circle. For example: the short way round.
subtend
to make an angle at a point from the two ends of a line. For example: the chord subtends 140° at the centre.
inscribed angle
an angle at the circumference between two chords. For example: angle APB in the figure.
alternate segment
the part of the circle on the other side of the chord from the tangent-chord angle. For example: the side the angle does not open towards.

Two surveyors and a pond wall

A chord AB across a round pond subtends 140° at the centre O. A first surveyor P walks round the long way, the major arc. A second surveyor Q walks round the short way, the minor arc. Each measures the angle between the lines of sight to A and B.

Predict first: what will P read, and what will Q read? Write your two numbers and your reason, then use the first figure to walk a point right round the circle.

90
70.0°angle APB
major arcwhich arc P is on

P at 90° is on the major arc: angle APB = 70.0°, half of the centre angle 140°.

Show me the steps
  1. P at 90°, on the major arc: angle APB = 70.0°, half of the centre angle AOB = 140°.
  2. Move P to 30° and then to 150°: still 70.0°. Every point of the major arc sees AB at the same angle: the same-segment theorem.
  3. Move P to 270°, on the minor arc: angle APB = 110.0°. This is not half of 140°; it is 180° − 70°, the opposite angle of a cyclic quadrilateral APBQ with Q on the major arc.
  4. Move P to 210° and 330°, both on the minor arc: 110.0° each. A point on the minor arc always reads 110.0° and a point on the major arc always 70.0°; the two readings add to 180°.

The angle at the centre is twice the angle at the circumference on the same arc. Write angle APB as x + y by joining PO. Case 1, O inside the angle: OP = OA = OB, so the two isosceles triangles have base angles x and y, and the angle at the centre is 2x + 2y. Case 2, O on PB: triangle OPA gives an exterior angle 2x at O, which is the whole centre angle. Case 3, O outside the angle: the centre angle is 2y − 2x, which is twice the angle y − x at P.

Consequences. Angles on the same arc are equal; the angle in a semicircle is 90°; opposite angles of a cyclic quadrilateral add to 180°, because the two centre angles on the two arcs make 360°; and the angle between a tangent and a chord equals the angle in the alternate segment.

240
50.0°angle APB
O is inside the anglewhere the centre is

P at 240°: O is inside the angle, and APB = 50.0°, half of AOB = 100°.

Show me the steps
  1. P at 240°: O lies inside the angle APB. Extending PO splits the angle into x + y, and the centre angle is 2x + 2y: the first case of the proof.
  2. P at 180°: PA is a diameter, so O lies on the line PA, a side of the angle. Triangle OPB is isosceles and the exterior angle at O is 2 × the angle at P: the second case.
  3. P at 120°: O is outside the angle APB. The angle at the centre is 2y − 2x for the two isosceles triangles, so it is still 2 × (y − x) = 2 × angle APB: the third case.
  4. In all three, the angle APB is 50.0°. A proof for one case alone would leave the others open; the three together settle every position of P on the major arc.

“Half the centre angle” belongs to the same arc. A point on the minor arc sees the chord from the other side. It reads 180° minus half the centre angle, not half of it. Say which arc P is on before you halve.

Worked examples

1. Prove the third case: O outside the angle APB, with the ray PA between PO and PB.

  1. Extend PO to meet the circle again at Q. Let angle OPA = x and angle OPB = y, so y is larger than x and angle APB = y − x.
  2. OP = OA gives angle OAP = x, and OP = OB gives angle OBP = y.
  3. Angle AOQ = 2x and angle BOQ = 2y, each an exterior angle of an isosceles triangle.
  4. Angle AOB = angle BOQ − angle AOQ = 2y − 2x = 2(y − x) = 2 × angle APB.

2. Points A, B, C, D lie on a circle in that order and O is the centre. Angle ABC = 112°. Find the reflex angle AOC and the angle ADC.

  1. B is on the minor arc AC, so the centre angle on the arc ADC, the other way, is the reflex angle AOC = 2 × 112° = 224°.
  2. Angle ADC is on the other arc, so it is half of the centre angle on arc ABC: (360° − 224°) ÷ 2 = 68°. Check: 112° + 68° = 180°.

3. Prove that the angle between a tangent and a chord equals the angle in the alternate segment, for an acute angle.

  1. Let the tangent at T be TS and the chord TA, with angle STA = θ, acute. Draw the diameter TD. Then angle STD = 90°, because a tangent is perpendicular to the radius.
  2. Since θ < 90°, the ray TA lies between TS and TD, so S and D are on opposite sides of TA, and D is in the alternate segment.
  3. Angle TAD = 90°, the angle in a semicircle. Angle ATD = 90° − θ, so in triangle TAD, angle ADT = θ.
  4. Every other point of the alternate segment sees TA at the same angle as D, by the same-segment theorem.

Your go. The first step is given. A, B, C, D lie on a circle in that order. The diagonals AC and BD meet at X, with angle ABD = 41° and angle BAC = 26°. Find angle DXC.

  1. Angle ACD = angle ABD = 41°, both on the chord AD from the same side.
  2. Angle XDC = angle BDC = angle BAC = 26°, both on the chord BC. Use triangle DXC.
Check the answer

Angle DXC = 180° − 41° − 26° = 113°. Stopping at 41° + 26° = 67° gives the sum of the two base angles, not the angle at X. Subtract it from 180°.

Check before you go on

1. The angle at the centre AOB of a circle is 130°. The point P is on the minor arc AB. What is the angle APB?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. O is the centre of a circle and A and B are on the circle. The angle AOB is 124°. Find the angle OAB.

3. Triangle ABC is inscribed in a circle with centre O. Angle BAC = 38° and angle ABC = 71°. Find the angle AOB.

Level 2 · Building

4. A, B, C and D lie in that order on a circle with centre O. Angle ADC is three times angle ABC. Find the reflex angle AOC.

5. TS is a tangent to a circle with centre O, at T, and TA is a chord. Angle STA = 64°. Find the angle TOA.

6. ABCD is a cyclic quadrilateral with AB parallel to DC. Angle ADC = 78°. Find the angle ABC.

Level 3 · Stretch

7. From a point P outside a circle with centre O, two tangents PA and PB touch the circle at A and B. Angle APB = 48°. C is a point on the major arc AB. Find the angle ACB.

8. A, B, C and D lie on a circle in that order. The diagonals AC and BD meet at X. Angle ABD = 45° and angle DXC = 97°. Find angle BAC.

Something to think about

In the first figure the chord subtends 140° at the centre. What chord would make the angle at every point of the major arc equal to 90°, and what is then special about the minor arc?

One way to think about it

A chord that subtends 180° at the centre: a diameter. Then every point on either arc sees it at 90°, and the two arcs are semicircles, so the major and minor arcs are the same length and the angle in a semicircle is 90° on both sides.

Exit ticket

9. O is the centre of a circle and A and B are on the circle. Angle OAB = 35°. The point P is on the minor arc AB. What is angle APB?