Year 11 Standard · Unit 2 Right-angled triangles and circle geometry · Lesson 8 of 8

Four points on a circle, and a tangent meeting a chord

Put all four corners of a quadrilateral on one circle and ask what that does to its angles. This lesson finds out and proves it, then proves what a tangent and a chord do, and uses both with algebra.

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 7: three angles of a quadrilateral are 80°, 90° and 100°. Find the fourth.
    Answer90°, because the angles sum to 360°.
  2. From Year 11: the angle at the centre of a circle is 112°. What is the angle at the circumference on the same arc?
    Answer56°, because it is half the centre angle.
  3. From Year 9: solve 4x + 12 = 100.
    Answerx = 22, because 4x = 88.
Words used in this lesson
cyclic quadrilateral
a four-sided shape whose corners all lie on one circle. For example: four posts round a round pond.
opposite angles
angles at corners that do not share a side. For example: A and C in ABCD.
exterior angle
an angle formed outside a shape by extending one side. For example: the angle at C after extending BC.
alternate segment
the segment of a circle on the other side of a chord. For example: the segment across the chord TA from the marked angle at T.
tangent-chord angle
the angle between a tangent and a chord at the point of contact. For example: the angle at T.
converse
the result with its cause and effect swapped. For example: if opposite angles add to 180, the corners are on a circle.

Four corners on a circle

A designer draws a quadrilateral ABCD whose four corners all lie on the same circle, like four posts round a circular pond. She measures angle A and finds it is 100°. She wonders what angle C, the corner opposite, can be.

Predict first: can angle C be 100° as well, or must it be 80°, or can it be anything at all? Write down your choice and a reason, then use the first figure to test it.

90
110
93.0° and 87.0°angles A and C
97.0° and 83.0°angles B and D

Arcs AB = 90°, BC = 76°, CD = 110°, DA = 84°: angle A = 93.0°, B = 97.0°, C = 87.0°, D = 83.0°.

Show me the steps
  1. Arcs 90°, 76°, 110°, 84°: angle A = 93.0°, C = 87.0°, B = 97.0°, D = 83.0°. Both pairs of opposite angles add to 180°.
  2. Make arc AB 50° and keep CD 110°: DA grows to 124°. A = 93.0° and C = 87.0°, but B = 117.0° and D = 63.0°. Both opposite pairs still add to 180°.
  3. Make arc CD 50° and arc AB 90°: DA is 144°. A = 63.0°, C = 117.0°, B = 97.0°, D = 83.0°. Each angle is half an arc, and each opposite pair still adds to 180°.
  4. Whatever the arcs, two opposite angles stand on arcs that together make the whole circle, so they are halves of 360°.

Opposite angles of a cyclic quadrilateral add to 180°. Proof: join O to B and D. The angle at the centre standing on arc BCD is 2 × angle A, and the angle at the centre standing on arc BAD is 2 × angle C. Those two centre angles make a full turn, 360°, so 2A + 2C = 360° and A + C = 180°. An exterior angle of a cyclic quadrilateral equals the interior opposite angle.

The alternate segment theorem. The angle between a tangent and a chord equals the angle in the alternate (other) segment. Proof, for an acute angle θ: the radius OT is perpendicular to the tangent, so in the isosceles triangle OTA each base angle is 90° − θ, and the centre angle TOA is 2θ. The angle at P on the other arc is half of it, θ. If θ = 90°, TA is a diameter and the angle in a semicircle is 90°. If θ is obtuse, the angle 180° − θ on the other side of the chord is acute, so a point Q in its alternate segment gives 180° − θ. A point P in the segment for θ then makes TPAQ a cyclic quadrilateral, so angle TPA = 180° − (180° − θ) = θ.

40
4
40.0°tangent and chord angle
40.0°angle TPA at P

The tangent and chord make 40.0°. At P in the alternate segment the angle TPA is also 40.0°, at every position of P.

Show me the steps
  1. Tangent-chord angle 40°, P at position 4: the angle TPA is 40.0°, the same.
  2. Move P to position 8: the angle TPA is still 40.0°. Every point of the alternate arc gives the same angle.
  3. Make the tangent-chord angle 70°: the angle TPA becomes 70.0°. It always matches the angle between the tangent and the chord.
  4. Make the angle 20°: TPA = 20.0°. The proof: centre angle TOA = 2 × 20° = 40° (isosceles triangle OTA, base angles 70°), and the angle at P is half of it.

Opposite, not next to each other. In a cyclic quadrilateral, angles at A and C add to 180°, and so do B and D. Angles at A and B do not. With the alternate segment, the angle is on the far side of the chord from the angle you were given, not on the same side.

Worked examples

1. Prove that opposite angles of a cyclic quadrilateral add to 180°.

  1. Let O be the centre. The angle at the centre standing on arc BCD is twice angle BAD (centre angle theorem).
  2. The angle at O on the other side, standing on arc BAD, is twice angle BCD (centre angle theorem). One of the two angles at O is reflex unless BD is a diameter; lesson 7 proves the theorem for a reflex angle too.
  3. The two angles at O together make a full turn: 2 × angle A + 2 × angle C = 360°.
  4. So angle A + angle C = 180°.

2. PQRS is a cyclic quadrilateral with angle P = 78° and angle Q = 104°. Find angles R and S.

  1. R is opposite P: R = 180° − 78° = 102°.
  2. S is opposite Q: S = 180° − 104° = 76°. Check: 78 + 104 + 102 + 76 = 360.

3. Prove that the angle between a tangent and a chord equals the angle in the alternate segment.

  1. Take θ acute (the key card covers the other cases). Let the angle between the tangent at T and the chord TA be θ. The radius OT is perpendicular to the tangent, so angle OTA = 90° − θ.
  2. OT = OA, so angle OAT = 90° − θ too, and the centre angle TOA = 180° − 2(90° − θ) = 2θ.
  3. The angle at P on the major arc is half the centre angle: θ.

Your go. The first step is given. In a cyclic quadrilateral ABCD, angle ABC = (4x − 6)° and angle ADC = (2x + 12)°. Find the angle ABC.

  1. B and D are opposite, so (4x − 6) + (2x + 12) = 180.
  2. Solve for x, then work out the angle ABC.
Check the answer

6x + 6 = 180, so x = 29. Angle ABC = 4 × 29 − 6 = 110° and ADC = 2 × 29 + 12 = 70°, which add to 180°.

Check before you go on

1. ABCD is a cyclic quadrilateral and the angle ABC is 72°. What is the angle ADC?

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. ABCD is a cyclic quadrilateral. The angle DAB is 95°. What is the angle BCD, in degrees?

3. A tangent touches a circle at T, and TA is a chord. The angle between the tangent and the chord TA is 54°. What is the angle TPA at a point P in the alternate segment, in degrees?

Level 2 · Building

4. ABCD is a cyclic quadrilateral. The angle at A is (2x + 10)° and the angle at C is (3x − 5)°. Find x.

5. ABCD is a cyclic quadrilateral. The side BC is extended to E. The angle DCE is 68°. What is the angle DAB, in degrees?

6. A tangent at A touches a circle through A, B and C. The angle between the tangent and the chord AB, on the side away from C, is 48°, and the angle ABC is 65°. What is the angle BAC, in degrees?

Level 3 · Stretch

7. ABCD is a cyclic quadrilateral and AB is a diameter. The angle CAB is 30°. What is the angle ADC, in degrees?

8. A tangent at T and a chord TA make an angle of (3x + 4)°. The angle TPA in the alternate segment is (5x − 22)°. What is the size of the angle between the tangent and the chord, in degrees?

Something to think about

Can a parallelogram that is not a rectangle be cyclic? Use the opposite-angle result to decide, and say what shape a cyclic parallelogram must be.

One way to think about it

No: opposite angles of a parallelogram are equal, and in a cyclic quadrilateral they add to 180°, so each is 90°. A cyclic parallelogram has four right angles, so it is a rectangle.

Exit ticket

9. ABCD is a cyclic quadrilateral. The diagonal BD splits the angle ADC into angle ADB = 34° and angle BDC = 52°. What is the angle ABC?