Chapter 26 · Higher tier

GCSE Maths Circle Theorems (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use the words chord, tangent, arc, sector and segment correctly, and spot isosceles triangles made from radii
Know, use and prove that the angle in a semicircle is 90°
Know, use and prove that the angle at the centre is twice the angle at the circumference
Know and use that angles in the same segment are equal, and that opposite angles of a cyclic quadrilateral add up to 180°
Know and use that a tangent is perpendicular to the radius, and that two tangents from a point are equal in length
Know, use and prove the alternate segment theorem
Know and use that the perpendicular from the centre to a chord bisects the chord, with Pythagoras and trigonometry
Combine theorems with isosceles triangles and algebra, give reasons in standard wording, and write short proofs

You should already know

Angle facts, isosceles triangles and angle sums (Chapter 17)
Circles: radius, diameter, circumference (Chapter 18)
Pythagoras' theorem and basic trigonometry (Chapter 20)
Solving linear equations (Chapter 9)

Where this leads

Equation of a circle and the tangent to a circle (Chapter 25)
Geometric proof in vector geometry (Chapter 22)
Further circle geometry in A-level mathematics

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: G9, G10.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

26.1 Circle words and isosceles triangles: Worksheet (PDF) · Answers (PDF) · lesson page
26.2 The angle in a semicircle: Worksheet (PDF) · Answers (PDF) · lesson page
26.3 The angle at the centre: Worksheet (PDF) · Answers (PDF) · lesson page
26.4 Angles in the same segment: Worksheet (PDF) · Answers (PDF) · lesson page
26.5 Cyclic quadrilaterals: Worksheet (PDF) · Answers (PDF) · lesson page
26.6 Tangents and radii: Worksheet (PDF) · Answers (PDF) · lesson page
26.7 The alternate segment theorem: Worksheet (PDF) · Answers (PDF) · lesson page
26.8 Chords: Worksheet (PDF) · Answers (PDF) · lesson page
26.9 Problems, algebra and proof: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FTwo angles of a triangle are 38° and 77°. Work out the third angle.

Done

FAn isosceles triangle has an angle of 40° between its two equal sides. Work out the other two angles.

Done

FAngles on a straight line: one angle is 118°. Work out the other angle on the line.

Done

FA right-angled triangle has shorter sides 9 cm and 12 cm. Work out the length of the hypotenuse.

Done

FSolve 5x − 12 = 3x + 20.

Done

26.1

Circle words and isosceles triangles

Edexcel 1MA1: G9, G10
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Circle words and isosceles trianglesComing soon

A circle is made of all the points that are the same distance from one fixed point, the centre. You met the main words in Chapter 18. Circle theorems depend on them, so every word must be exact. A diagram usually uses lots of these words at once, and you need to spot them quickly.

Chord AB (the blue region above it is a segment). CD passes through the centre O, so it is a diameter. The orange region OEF is a sector, and the curved line from E to F is an arc. The red line HJ touches the circle at G: it is a tangent.

Words to know

Radius: a straight line from the centre to a point on the circle. All radii of one circle are equal.
Diameter: a chord through the centre. It is twice the radius, and it is the longest chord.
Chord: a straight line that joins two points on the circle.
Arc: a part of the circumference. The shorter one is the minor arc, the longer is the major arc.
Sector: the region between two radii and an arc (a slice of pizza).
Segment: the region between a chord and an arc. A chord cuts a circle into two segments.
Tangent: a straight line that touches the circle at exactly one point, the point of contact.

Two radii always make an isosceles triangle. Every radius is the same length. So if A and B are on a circle with centre O, then OA = OB and triangle OAB is isosceles. The two equal angles (the base angles) are at A and B, which are the points on the circle. This one idea is used in almost every proof in the chapter. Remember from Chapter 17: in an isosceles triangle the angles opposite the equal sides are equal, and the three angles add up to 180°.

Triangle OAB is isosceles because OA and OB are radii (the tick marks show equal lengths). The two base angles are equal, here 52° each.

Writing reasons. In the exam, circle theorem questions say “give reasons for your answer”. You get marks for the angle and for the reason, and the reason must be in the right words. Learn the wording in the box below and use it exactly. Each section of this chapter shows you how a theorem works and why it is true.

The reasons you must be able to write

Angle in a semicircle is 90°.
The angle at the centre is twice the angle at the circumference.
Angles in the same segment are equal.
Opposite angles of a cyclic quadrilateral add up to 180°.
The angle between a tangent and a radius is 90°.
Tangents from an external point are equal in length.
The angle between a tangent and a chord is equal to the angle in the alternate segment.
The perpendicular from the centre to a chord bisects the chord.
Also use: base angles of an isosceles triangle are equal; angles in a triangle add up to 180°; angles on a straight line add up to 180°; angles in a quadrilateral add up to 360°.

Watch out

A chord is not the same as a radius. A radius has one end at the centre. A chord has both ends on the circle. The only chord that passes through the centre is the diameter.

Worked example 26.1.1

The diagram shows a circle with centre O. Name (a) the line CD, (b) the shaded region OEF, (c) the shaded region next to the line AB, (d) the line HJ, which touches the circle at G.

Method. Match each description to the exact circle word.

Look at CD. It joins two points on the circle and passes through the centre O.
(a) CD is a diameter (a chord through the centre)
Look at the region OEF. It is bounded by two radii (OE and OF) and an arc.
(b) a sector
Look at the region by AB. AB is a chord that does not pass through the centre. The region between the chord and its arc is cut off from the rest of the circle.
(c) a segment
Look at HJ. It touches the circle at one point, G, and does not cross into the circle.
(d) a tangent

Check. A sector has two radii as straight sides. A segment has a chord as its straight side. ✓

Watch out. A sector is bounded by radii; a segment is bounded by a chord. Both have an arc.

Worked example 26.1.2

O is the centre of the circle. A and B are points on the circle. Angle AOB = 70°. Work out the size of angle OAB.

Key words. A radius joins the centre to the circle. A chord joins two points on the circle. Isosceles means two equal sides, which gives two equal angles.

Method. Two radii of a circle are equal, so any triangle made from two radii and a chord is isosceles. Then use the isosceles and angle-sum facts.

Spot the radii. Every line from the centre to the circle is a radius, and all radii of one circle are the same length.
OA and OB are both radii, so OA = OB.
Name the triangle. A triangle with two equal sides is isosceles.
Triangle OAB is isosceles, with the two equal sides meeting at O.
Use the angle sum. The three angles of any triangle add up to 180°. The two base angles (at A and B) are equal because the triangle is isosceles.
Angle OAB + angle OBA + angle AOB = 180°
2 × angle OAB + 70° = 180°
Solve for the base angle. Take away the known angle, then share the rest equally between the two base angles.
2 × angle OAB = 180° − 70° = 110°
angle OAB = 110° ÷ 2 = 55°

Check. The three angles are 55° + 55° + 70° = 180°, which is a full triangle. ✓

Watch out. Do not give the base angles as the angle at the centre. In triangle OAB the equal angles are the two at the circumference, A and B.

Worked example 26.1.3

O is the centre of the circle. A and B are points on the circle. Angle OAB = 25°. Work out the size of angle AOB.

Key words. A radius joins the centre to the circle. A chord joins two points on the circle. Isosceles means two equal sides, which gives two equal angles.

Method. Two radii of a circle are equal, so any triangle made from two radii and a chord is isosceles. Then use the isosceles and angle-sum facts.

Spot the radii. All radii of a circle are the same length.
OA and OB are both radii, so OA = OB and triangle OAB is isosceles.
Write down the other base angle. The base angles of an isosceles triangle are equal.
Angle OBA = angle OAB = 25°
Use the angle sum. The three angles of a triangle add up to 180°, so subtract the two base angles from 180°.
Angle AOB = 180° − 25° − 25° = 130°

Check. 25° + 25° + 130° = 180°. ✓

Watch out. Do not give the base angles as the angle at the centre. In triangle OAB the equal angles are the two at the circumference, A and B.

Worked example 26.1.4

A and B are points on a circle with centre O. The length of the chord AB is equal to the radius of the circle. Show that angle AOB = 60°.

Method. Show that all three sides of triangle OAB are equal. Then the triangle is equilateral and each angle is 60°.

List the three sides. OA and OB are radii of the circle. AB is the chord, and the question says it is equal to the radius.
OA = OB = radius, and AB = radius
Say what kind of triangle it is. A triangle with three equal sides is equilateral.
Triangle OAB is equilateral.
Use the angle fact. The three angles of a triangle add up to 180°, and in an equilateral triangle they are all equal.
180° ÷ 3 = 60°
Write the conclusion. End a “show that” question with a clear statement.
So angle AOB = 60°. ✓

Check. Another way: the base angles are equal and so is the apex, so 3 × 60° = 180°. ✓

Watch out. In a “show that” question, state each fact and reason; do not just draw a number from nowhere.

26.2

The angle in a semicircle

Edexcel 1MA1: G10
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The angle in a semicircleComing soon

Draw any diameter AB of a circle, then pick any point C on the circle and join it to A and to B. Whichever point C you choose, the angle at C is always a right angle.

Angle ACB = 90° wherever C is on the circle (except at A or B), because AB is a diameter.

This is the angle in a semicircle theorem. The name comes from the diameter AB cutting the circle into two semicircles, and C being a point on the curved edge of one of them. Reason to write: “angle in a semicircle is 90°”. It is one of the most useful facts: a right angle lets you use Pythagoras' theorem and trigonometry (Chapter 20).

Why is it true? Join C to the centre O. Now OA, OB and OC are all radii, so they are equal. That makes two isosceles triangles, OAC and OBC. Call the equal base angles a (in triangle OAC) and b (in triangle OBC). The angle at C in the big triangle is a + b. The angles of triangle ABC add up to 180°, so a + b + (a + b) = 180°, which means 2(a + b) = 180° and a + b = 90°. You will see this written out as a proof in worked example 26.2.4.

The proof: OA, OC and OB are equal radii. The angles marked a are equal, and the angles marked b are equal.

Key ideas

If AB is a diameter and C is on the circle, then angle ACB = 90°.
Reason: “angle in a semicircle is 90°”.
The right angle is at the point on the circle, opposite the diameter.
The right-angled triangle ABC has the diameter as its hypotenuse (longest side).
The converse is also true: if a triangle with a right angle at C is drawn in a circle through A, B and C, then AB is a diameter.

Watch out

The theorem only works when the line is a diameter. If the line AB does not pass through the centre, the angle at C is not 90°. Always check for the centre O (or the word “diameter”) before you use it.

Worked example 26.2.1

AB is a diameter of the circle. C is a point on the circle. Angle CAB = 35°. Work out the size of angle ABC. Give a reason for each step.

Key words. A diameter is a chord through the centre. An angle in a semicircle is the angle made at the circle by lines drawn to both ends of a diameter.

Method. A diameter makes a right angle with every point on the circle. Use that right angle, then the angle sum of a triangle.

Mark the triangle ABC. The diameter AB and the point C on the circle make a triangle. We are looking for angle ABC in it.
Angle CAB = 35°, angle ABC = ?
Find angle ACB. The line AB is a diameter, so the angle at C on the circle is an angle in a semicircle.
Angle ACB = 90° (reason: angle in a semicircle is 90°)
Use the angle sum of the triangle. The three angles of triangle ABC add up to 180°, so subtract the two we know.
Angle ABC = 180° − 90° − 35°
Angle ABC = 55° (reason: the angles in a triangle add up to 180°)

Check. 35° + 55° + 90° = 180°. ✓

Watch out. The right angle is at the point on the circle (C), not at the end of the diameter.

Worked example 26.2.2

AB is a diameter of the circle with centre O. C is a point on the circle. Angle ABC = 66°. Work out the size of angle ACO. Give a reason for each step.

Key words. A diameter is a chord through the centre. An angle in a semicircle is the angle made at the circle by lines drawn to both ends of a diameter.

Method. A diameter makes a right angle with every point on the circle. Use that right angle, then the angle sum of a triangle.

Find angle ACB. The line AB is a diameter, so the angle at C is in a semicircle.
Angle ACB = 90° (reason: angle in a semicircle is 90°)
Find angle OCB. OC and OB are both radii, so triangle OCB is isosceles. Its base angles are equal.
Angle OCB = angle OBC = 66° (reason: the base angles of an isosceles triangle are equal)
Subtract to get angle ACO. Angle ACB is made of angle ACO and angle OCB side by side.
Angle ACO = 90° − 66° = 24°

Check. Triangle OAC is also isosceles (OA = OC) with angle OAC = 24°, so angle ACO = 24°. ✓

Watch out. The right angle is at the point on the circle (C), not at the end of the diameter.

Worked example 26.2.3

AB is a diameter of the circle. Angle CAB = (3x + 10)° and angle ABC = (3x + 20)°. Work out the value of x.

Method. The angle in a semicircle is 90°, so the other two angles of the triangle add up to 90°. That gives an equation in x.

Find angle ACB. The line AB is a diameter, so angle ACB is an angle in a semicircle.
Angle ACB = 90° (angle in a semicircle is 90°)
Write an equation. The three angles of triangle ABC add up to 180°. With the right angle taken away, the other two must add to 90°.
(3x + 10) + (3x + 20) = 90
Collect the like terms. Add the x terms together and the numbers together.
6x + 30 = 90
Subtract 30 from both sides. We want the number terms on the other side.
6x = 90 − 30
6x = 60
Divide both sides by 6. 6x means 6 × x, so dividing by 6 leaves x on its own.
x = 60 ÷ 6 = 10
Check the answer by working out both angles. Put x back into each angle. They should add up to 90°.
3 × 10 + 10 = 40° and 3 × 10 + 20 = 50°
40° + 50° = 90° ✓

Check. The right angle plus 40° plus 50° is 90° + 90° = 180°. ✓

Watch out. The equation is "angles add to 90°", not 180°, because the right angle is already used up.

Worked example 26.2.4

AB is a diameter of the circle. AB = 17 cm and AC = 8 cm. Work out the length of BC.

Key words. The hypotenuse is the longest side of a right-angled triangle, opposite the right angle.

Method. A diameter gives a right angle at the circle, so Pythagoras' theorem can be used in triangle ABC.

Find the right angle. AB is a diameter, so the angle at C is an angle in a semicircle.
Angle ACB = 90° (angle in a semicircle is 90°)
Write Pythagoras for triangle ABC. The longest side (the hypotenuse) is opposite the right angle. It is AB, the diameter.
AB² = AC² + BC²
17² = 8² + BC²
Work out the squares.
289 = 64 + BC²
Rearrange for BC². Subtract the known square from both sides.
BC² = 289 − 64 = 225
Square root. Lengths are positive, so take the positive root.
BC = √225 = 15 cm

Check. 8² + 15² = 64 + 225 = 289 = 17². ✓

Watch out. Pythagoras: the diameter is the longest side (hypotenuse). Never add the squares of the hypotenuse.

Worked example 26.2.5

Prove that the angle in a semicircle is a right angle. In the diagram AB is a diameter of a circle with centre O, and C is a point on the circle.

Method. Join C to the centre O to make two isosceles triangles, then use the angle sum of triangle ABC.

Draw the radius OC. This splits the triangle into two smaller triangles that we can reason about.
OA = OC = OB (all radii of the circle)
Label the base angles in triangle OAC. It is isosceles because OA = OC, so its base angles are equal. Call each one a.
angle OAC = angle OCA = a
Label the base angles in triangle OBC. It is isosceles because OB = OC. Call the base angles b.
angle OBC = angle OCB = b
Write angle ACB. It is made of the two angles at C side by side.
angle ACB = a + b
Use the angle sum in triangle ABC. The angles of a triangle add up to 180°. The three angles are a at A, b at B and a + b at C.
a + b + (a + b) = 180°
2(a + b) = 180°
Divide by 2 and conclude. Divide both sides by 2.
a + b = 90°, so angle ACB = 90°. The angle in a semicircle is a right angle. ✓

Check. Try numbers: if a = 25° and b = 65° then angle ACB = 90°, and the triangle angles are 25°, 65° and 90°, which add up to 180°. ✓

Watch out. A proof must work for every position of C, so use letters (a and b) instead of numbers.

26.3

The angle at the centre

Edexcel 1MA1: G10
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The angle at the centreComing soon

Look at an arc AB of a circle. Two kinds of angle can be drawn standing on that arc, which means their arms end at A and B. One has its corner at the centre O. The other has its corner at a point C on the circumference (the rest of the circle). The theorem says the angle at the centre is exactly twice the angle at the circumference.

Both angles stand on arc AB. Angle AOB (at the centre) = 2 × angle ACB (at the circumference).

The reason to write is “the angle at the centre is twice the angle at the circumference”. To use it, always do this first: name the arc both angles stand on. If you pick the wrong arc, you will double or halve the wrong angle. Then double to go from the circumference to the centre, and halve to go from the centre to the circumference.

Why is it true? Join C to O and carry on to the point D on the other side of the circle. OA, OB and OC are radii, so triangle OAC is isosceles with base angles p. Its angle at O is 180° − 2p. Angles on the straight line CD add to 180°, so angle AOD = 2p. In the same way triangle OBC has base angles q and angle BOD = 2q. So angle AOB = 2p + 2q = 2(p + q) = 2 × angle ACB. The full proof is worked example 26.3.5.

The proof: join C to O and extend to D. Each isosceles triangle gives a pair of equal base angles.

Special cases. (1) If the centre O lies on the line AB (a diameter) the angle at the centre is 180°, which gives the angle in a semicircle: 90°. (2) If C is on the minor arc AB, then the angle at the circumference stands on the major arc, so it matches the reflex angle at the centre (the one bigger than 180°). Find the reflex angle with 360° − the angle shown, then halve it. Worked example 26.3.3 shows this.

Key ideas

The angle at the centre is twice the angle at the circumference, when both stand on the same arc.
Reason: “the angle at the centre is twice the angle at the circumference”.
Circumference to centre: double. Centre to circumference: halve.
If the point is on the minor arc, use the reflex angle at the centre: 360° minus the angle shown.
Radii OA and OB make an isosceles triangle, so the centre theorem often combines with base angles.

Watch out

Both angles must stand on the same arc. Do not double an angle that is not made from the same two points A and B. And remember: the angle at the circumference is the smaller one (unless the point is on the minor arc).

Worked example 26.3.1

A, B and C are points on a circle with centre O. Angle ACB = 42°. Work out the size of angle AOB. Give a reason for your answer.

Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.

Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.

Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.
Centre: angle AOB = ?
Circumference: angle ACB = 42°
Choose the rule. We are going from the circumference to the centre, so we use the circle theorem for the angle at the centre.
angle at the centre = 2 × angle at the circumference
Double it. The angle at the centre is twice the angle at the circumference.
Angle AOB = 2 × 42° = 84°

Check. Half of 84° is 42°. ✓

Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.

Worked example 26.3.2

A, B and C are points on a circle with centre O. Angle ACB = 50°. Work out the size of angle OAB. Give a reason for each step.

Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.

Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.

Find angle AOB. Angle at the centre and angle at the circumference are both on the same arc AB.
Angle AOB = 2 × 50° = 100° (reason: the angle at the centre is twice the angle at the circumference)
Spot the isosceles triangle. OA and OB are radii, so they are equal. Triangle OAB is isosceles and its base angles are equal.
Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
Use the angle sum in triangle OAB. The angles add up to 180°. Take off the angle at O and share the rest equally.
2 × angle OAB = 180° − 100° = 80°
Angle OAB = 80° ÷ 2 = 40°

Check. 40° + 40° + 100° = 180°. ✓ Also 40 = 90 − 50. ✓

Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.

Worked example 26.3.3

A, B and C are points on a circle with centre O. C lies on the minor arc AB. Angle AOB = 140°. Work out the size of angle ACB. Give a reason for each step.

Key words. The minor arc is the shorter arc between two points and the major arc is the longer one. A reflex angle is bigger than 180°.

Method. If the point is on the minor arc, the angle at the circumference stands on the major arc, so we use the reflex angle at the centre.

Decide which arc the angle stands on. C is on the minor arc AB, so angle ACB looks at the chord AB from the minor-arc side. It stands on the other arc, the major arc.
angle ACB stands on the major arc AB
Find the reflex angle at the centre. Angle ACB stands on the major arc AB (the long way round, the arc that does not contain C). The matching angle at the centre is the reflex angle AOB.
Reflex angle AOB = 360° − 140° = 220° (angles around a point add up to 360°)
Halve it. The angle at the centre is twice the angle at the circumference.
Angle ACB = 220° ÷ 2 = 110° (the angle at the centre is twice the angle at the circumference)

Check. Using the cyclic quadrilateral idea: 110° + 70° = 180°, where 70° is the angle at the circumference from the major arc. ✓

Watch out. An angle at the circumference on the minor arc is obtuse. Do not just halve the angle shown at O; halve the reflex angle.

Worked example 26.3.4

A, B and C are points on a circle with centre O. Angle AOB = (5x + 12)° and angle ACB = (2x + 12)°.

Work out the value of x.
Work out the size of angle ACB.

Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.

Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.
(5x + 12) = 2 × (2x + 12) (the angle at the centre is twice the angle at the circumference)
Expand the bracket on the right. Multiply both terms inside by 2.
2 × (2x + 12) = 4x + 24
Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.
5x + 12 = 4x + 24
Subtract 4x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.
5x + 12 − 4x = 4x + 24 − 4x
x + 12 = 24
Subtract 12 from both sides. We want the number terms on the other side.
x = 24 − 12
x = 12
Read off x.
x = 12
Work out angle ACB. Put x into the expression for the angle.
Angle ACB = 2 × 12 + 12 = 36°

Check. Angle AOB = 5 × 12 + 12 = 72° and 2 × 36° = 72°. ✓

Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.

Worked example 26.3.5

Prove that the angle at the centre of a circle is twice the angle at the circumference. A, B and C are on a circle with centre O. Show that angle AOB = 2 × angle ACB.

Method. Join C to O and carry on to D. Use the isosceles triangles OAC and OBC and angles on a straight line.

Draw the line COD. It passes through the centre. We can now split the problem into two halves, one for A and one for B.
OA = OB = OC (radii)
Triangle OAC. OA = OC, so the triangle is isosceles and its base angles are equal. Call each one p.
angle OAC = angle OCA = p, so angle AOC = 180° − 2p
Find angle AOD. C, O and D are in a straight line, so angle AOC and angle AOD add up to 180°.
angle AOD = 180° − (180° − 2p) = 2p
Do the same for triangle OBC. OB = OC, so the base angles are equal. Call each one q.
angle OBC = angle OCB = q, and in the same way angle BOD = 2q
Add the two angles at the centre. Angle AOB is made of angle AOD and angle DOB side by side.
angle AOB = 2p + 2q = 2(p + q)
Compare with the angle at C. Angle ACB is made of angle ACO and angle OCB.
angle ACB = p + q, so angle AOB = 2 × angle ACB. ✓

Check. Test: if p = 30° and q = 20° then angle ACB = 50° and angle AOB = 2 × 30 + 2 × 20 = 100°, which is twice 50°. ✓

Watch out. The proof shown is for the case where O is inside angle ACB. The other positions of O work the same way.

26.4

Angles in the same segment

Edexcel 1MA1: G10
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Angles in the same segmentComing soon

A chord cuts a circle into two segments (Section 26.1). Take any chord AB. Pick two points C and D on the circumference that are on the same side of the chord, in the same segment. The angle ACB and the angle ADB are equal. It does not matter where in the segment you put the points: the angle made at the circumference by the chord stays the same.

Both marked angles are made from the chord AB and the corners C and D are in the same segment. So the two angles are equal.

The reason to write is “angles in the same segment are equal”. Why is it true? Each of the two angles is half of the angle at the centre on the same arc AB (the centre theorem from 26.3). Two numbers that are both half of the same angle must be equal. The proof is worked example 26.4.5.

How to spot a pair. (1) Find the chord both angles share: it joins the other end of each arm. (2) Check the two corners are on the same side of the chord. In a picture with two chords crossing, like the one in the next examples, you can find several pairs: for each side of a quadrilateral ABCD, the two other corners see it at equal angles. If the two corners are on opposite sides of the chord, the angles are not equal: they add up to 180°, which is the subject of the next section.

Key ideas

Angles made by the same chord, with their corners on the same side of it (the same segment), are equal.
Reason: “angles in the same segment are equal”.
To find a pair: name the chord, then pick another corner on the same side of it.
Each angle is half of the angle at the centre on the same arc.
In a cyclic quadrilateral ABCD with diagonals AC and BD: BAC = BDC, ABD = ACD, CAD = CBD, ACB = ADB.

Watch out

The two angles must be made from the same chord. Angle ACB and angle ACD share one arm but are made from different chords, so they are not equal in general.

Worked example 26.4.1

A, B, C and D are points on a circle. Angle ACB = 50°. Work out the size of angle ADB. Give a reason for your answer.

Key words. A chord cuts a circle into two segments. Two angles are in the same segment if their corners are on the same side of the chord.

Method. Two angles made from the same chord, with their corners on the same side of it, are in the same segment and are equal.

Find the chord that both angles are drawn from. Both angles have their arms ending at A and B. So both are made from the chord AB.
Angle ACB and angle ADB both stand on chord AB.
Check the two points are on the same side of the chord. C and D are both on the same side of the line AB, so they lie in the same segment.
C and D are in the same segment of the circle.
Use the theorem. Angles in the same segment are equal.
Angle ADB = angle ACB = 50°

Check. Angle at the centre on the same arc would be 2 × 50° = 100° for either point. ✓

Watch out. Make sure both corners are on the same side of the chord. Angles on opposite sides of the chord add up to 180°, not equal.

Worked example 26.4.2

A, B, C and D are points on a circle. Angle ACB = (3x + 5)° and angle ADB = (2x + 20)°.

Work out the value of x.
Work out the size of angle ACB.

Method. Angles in the same segment are equal, so the two expressions are equal. Solve for x, then substitute.

Spot that the two angles are equal. Both angles are made from the chord AB with the corners C and D on the same side of it. They are in the same segment.
Angle ACB = angle ADB (angles in the same segment are equal)
Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.
3x + 5 = 2x + 20
Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.
3x + 5 − 2x = 2x + 20 − 2x
x + 5 = 20
Subtract 5 from both sides. We want the number terms on the other side.
x = 20 − 5
x = 15
Read off x.
x = 15
Find the angle. Put x into the expression.
Angle ACB = 3 × 15 + 5 = 50°

Check. Other angle: 2 × 15 + 20 = 50°. They match. ✓

Watch out. Equal angles means the expressions are equal (set one equal to the other), not that they add up to 180°.

Worked example 26.4.3

A, B, C and D are points on a circle. The chords AC and BD meet at E. Angle BAC = 35° and angle AEB = 105°. Work out the size of angle ACD. Give a reason for each step.

Method. Use the triangle angle sum first to find an angle that is in the same segment as the one we want.

Pick out triangle ABE. The chords cross at E. Triangle ABE contains the angle BAC and the angle AEB that we are given.
triangle ABE: angle BAE = 35°, angle AEB = 105°
Find angle ABE (that is, angle ABD). Triangle ABE has angle sum 180°. We know two of its angles.
Angle ABD = 180° − 35° − 105° = 40° (the angles in a triangle add up to 180°)
Match it with an angle in the same segment. Angles ABD and ACD are both made from the chord AD, with B and C on the same side of it.
Angle ACD = angle ABD = 40° (angles in the same segment are equal)

Check. In triangle DEC the angles are 35° (angle BDC, same segment as BAC), 40° and 105° (vertically opposite AEB): 35 + 40 + 105 = 180. ✓

Watch out. A "same segment" pair needs the same chord. Name the chord before you say the angles are equal.

Worked example 26.4.4

A, B, C and D are points on a circle. The chords AC and BD meet at E. Angle BAC = 40° and angle ACD = 45°. Work out the size of angle AEB. Give a reason for each step.

Method. Use the same-segment theorem to find a second angle in triangle DEC, then the angle sum and vertically opposite angles.

Find angle BDC. Angles BAC and BDC are both made from the chord BC, with A and D on the same side of it.
Angle BDC = angle BAC = 40° (angles in the same segment are equal)
Use triangle DEC. Its angles at D and C are now known: angle EDC is angle BDC and angle ECD is angle ACD. The angles add up to 180°.
Angle DEC = 180° − 40° − 45° = 95° (the angles in a triangle add up to 180°)
Use vertically opposite angles. The chords cross at E, so angle AEB is opposite angle DEC.
Angle AEB = angle DEC = 95° (vertically opposite angles are equal)

Check. Second route: angle ABD = angle ACD = 45° (same segment, chord AD). In triangle ABE: 180° − 40° − 45° = 95°. ✓

Watch out. Do not forget vertically opposite angles at the point where two chords cross.

Worked example 26.4.5

Prove that angles in the same segment are equal. A, B, C and D are on a circle, with C and D on the same side of the chord AB. Show that angle ACB = angle ADB.

Method. Use the centre theorem to show that each angle is half of the same centre angle.

Draw the angle at the centre. Join A and B to the centre O. Angle AOB stands on the same arc AB as both angles.
Let angle AOB = 2x
Use the centre theorem on C. The angle at the centre is twice the angle at the circumference, so the angle at the circumference is half the angle at the centre.
angle ACB = 2x ÷ 2 = x
Use the centre theorem on D. D is also on the circumference and its angle stands on the same arc AB.
angle ADB = 2x ÷ 2 = x
Compare. Both angles equal x.
angle ACB = angle ADB. Angles in the same segment are equal. ✓

Check. If angle AOB = 100° then both angles are 50°, wherever C and D are on the major arc. ✓

Watch out. This proof assumes C and D are on the major arc. For the minor arc, both angles are half of the reflex angle at the centre.

26.5

Cyclic quadrilaterals

Edexcel 1MA1: G10
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Cyclic quadrilateralsComing soon

A cyclic quadrilateral is a four-sided shape with all four corners on a circle. The word “cyclic” comes from the Greek word for circle. The theorem says: opposite angles of a cyclic quadrilateral add up to 180°. Opposite means across the shape, not next to each other.

In cyclic quadrilateral ABCD, angle A + angle C = 180° and angle B + angle D = 180°.

The reason to write is “opposite angles of a cyclic quadrilateral add up to 180°”. Why is it true? Join B and D to the centre O. Let angle BAD = a. It stands on the arc BCD, so the angle at the centre on that arc is 2a (centre theorem). The angle round the other side of the centre is 360° − 2a (angles round a point). Angle BCD stands on that other arc, so it is half of that: (360° − 2a) ÷ 2 = 180° − a. Add: a + 180° − a = 180°. The full proof is worked example 26.5.5.

The proof: the angle at the centre is 2a on one side and 360° − 2a on the other.

Exterior angles. If you extend one side of a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle. This is a fast shortcut: the exterior angle and its neighbouring interior angle make a straight line (180°), and the interior angle plus its opposite angle make 180° too. So the exterior angle equals the opposite one. You can use the shortcut in the exam if you give the reason “exterior angle of a cyclic quadrilateral is equal to the interior opposite angle”, but it is safer to show the two steps.

Key ideas

Opposite angles of a cyclic quadrilateral add up to 180°.
Reason: “opposite angles of a cyclic quadrilateral add up to 180°”.
Opposite means A with C, and B with D. Next-door angles do not add to 180° in general.
The exterior angle (made by extending a side) equals the interior opposite angle.
The angles of every quadrilateral add up to 360°; a cyclic quadrilateral has the extra property above.

Watch out

Only opposite angles add up to 180°. And the theorem needs all four corners on the circle. A parallelogram that is not a rectangle has opposite angles that add to more or less than 180°, so it cannot be cyclic.

Worked example 26.5.1

ABCD is a cyclic quadrilateral. Angle DAB = 95°. Work out the size of angle BCD. Give a reason for your answer.

Key words. A cyclic quadrilateral is a four-sided shape with all four corners on a circle. Opposite angles are at corners that are not next to each other.

Method. In a cyclic quadrilateral, opposite angles add up to 180°. Subtract the known angle from 180°.

Check the angles are opposite. In a quadrilateral ABCD, A is opposite C and B is opposite D. Opposite corners are not next to each other.
Angle DAB and angle BCD are opposite corners.
Write the cyclic quadrilateral fact. All four corners are on the circle, so opposite angles add up to 180°.
Angle DAB + angle BCD = 180°
Subtract. Take the known angle away from 180°.
Angle BCD = 180° − 95° = 85°

Check. 95° + 85° = 180°. ✓

Watch out. Opposite means the corner across the quadrilateral. Neighbouring angles do not add up to 180°.

Worked example 26.5.2

ABCD is a cyclic quadrilateral. The side BC is extended to the point E. Angle DAB = 95°. Work out the size of angle ECD. Give a reason for each step.

Method. Find the interior angle at the corner first (opposite angles of a cyclic quadrilateral add to 180°), then use angles on a straight line.

Mark the exterior angle. The side BC has been extended to E. The angle ECD is outside the quadrilateral, between the extension and the next side CD.
We want angle ECD.
Find the interior angle at C. Corner C is opposite corner A in the cyclic quadrilateral, so these two angles add up to 180°.
Angle BCD = 180° − 95° = 85° (opposite angles of a cyclic quadrilateral add up to 180°)
Use the straight line. B, C and E are on a straight line, so angle BCD and angle ECD are on a straight line and add up to 180°.
Angle ECD = 180° − 85° = 95° (angles on a straight line add up to 180°)

Check. The exterior angle 95° equals the opposite interior angle 95°. This is a useful shortcut: the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. ✓

Watch out. The exterior angle is equal to the interior angle opposite the corner, not the one next to it.

Worked example 26.5.3

ABCD is a cyclic quadrilateral. Angle DAB = (5x + 20)° and angle BCD = (3x + 40)°.

Work out the value of x.
Work out angles DAB and BCD.

Method. Opposite angles of a cyclic quadrilateral add up to 180°. Write that as an equation, solve for x, then find the angles.

Write the angle fact. A and C are opposite corners of a cyclic quadrilateral, so the angles add up to 180°.
(5x + 20) + (3x + 40) = 180 (opposite angles of a cyclic quadrilateral add up to 180°)
Collect like terms. Add the x terms together, then the numbers.
8x + 60 = 180
Subtract 60 from both sides. We want the number terms on the other side.
8x = 180 − 60
8x = 120
Divide both sides by 8. 8x means 8 × x, so dividing by 8 leaves x on its own.
x = 120 ÷ 8 = 15
Work out the two angles. Put x back into each expression.
Angle DAB = 5 × 15 + 20 = 95°
Angle BCD = 3 × 15 + 40 = 85°

Check. 95° + 85° = 180°. ✓

Watch out. Put each expression in brackets and add them. Do not try to make one expression equal the other.

Worked example 26.5.4

A, B, C and D are points on a circle with centre O. Angle AOC = 160° (the angle at O inside the quadrilateral on the side of D). Give a reason for each answer.

Work out angle ABC
Work out angle ADC

Part (a)

Method. Angle ABC and angle AOC stand on the same arc AC (the arc through D), so use the angle-at-the-centre theorem.

Spot the two angles. Angle AOC is at the centre, angle ABC is at the circumference, and both stand on the arc AC that contains D.
Centre: 160°
Circumference: angle ABC = ?
Halve it. The angle at the centre is twice the angle at the circumference.
Angle ABC = 160° ÷ 2 = 80° (the angle at the centre is twice the angle at the circumference)

Check. 2 × 80° = 160°. ✓

Watch out. Check which arc the angles stand on. B is on the other side of the circle from the arc ADC.

Part (b)

Method. ABCD is a cyclic quadrilateral, so opposite angles add up to 180°. Use the answer to part (a).

Spot the opposite corner. D is opposite B in the quadrilateral ABCD.
Angle ABC = 80° (from part a)
Subtract from 180°. Opposite angles of a cyclic quadrilateral add up to 180°.
Angle ADC = 180° − 80° = 100° (opposite angles of a cyclic quadrilateral add up to 180°)

Check. Second method: the reflex angle AOC = 360° − 160° = 200°, and half of it is 100°. ✓

Watch out. Angle ADC is obtuse here, because D is on the minor arc.

Worked example 26.5.5

Prove that opposite angles of a cyclic quadrilateral add up to 180°. ABCD is a quadrilateral with all four corners on a circle with centre O. Show that angle BAD + angle BCD = 180°.

Method. Join B and D to the centre. Use the centre theorem for each of the two angles, one on each arc.

Name the angle at A. Use a letter so the proof works for any size of angle.
Let angle BAD = a
Find the angle at the centre on the arc BCD. Angle BAD stands on arc BCD. The angle at the centre is twice the angle at the circumference.
angle BOD (on the C side) = 2a
Find the other angle at the centre. The full turn round O is 360°, so the angle on the other side is what remains.
reflex angle BOD = 360° − 2a (angles round a point add up to 360°)
Use the centre theorem again at C. Angle BCD stands on the arc BAD that goes the other way round, so it is half the reflex angle.
angle BCD = (360° − 2a) ÷ 2 = 180° − a
Add the opposite angles.
angle BAD + angle BCD = a + 180° − a = 180° ✓

Check. Try a = 80°: the angle at the centre is 160°, the reflex angle is 200°, angle BCD = 100°. 80° + 100° = 180°. ✓

Watch out. The same argument works for the other pair of opposite angles, B and D.

26.6

Tangents and radii

Edexcel 1MA1: G10
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Tangents and radiiComing soon

A tangent is a straight line that touches a circle at exactly one point, called the point of contact. It never goes inside the circle. There are two important facts about tangents.

Fact 1: a tangent is perpendicular to the radius at the point of contact. The angle between a tangent and the radius that meets it is 90°. Reason to write: “the angle between a tangent and a radius is 90°”. Why? Every point on the tangent except the point of contact is outside the circle, so it is further from the centre O than a radius. So the radius OT is the shortest distance from O to the tangent line. The shortest distance from a point to a straight line is always the perpendicular one. So OT is perpendicular to the tangent.

The tangent PQ touches the circle at T. The radius OT makes a right angle with it.

Fact 2: tangents from an external point are equal in length. If you draw two tangents TA and TB to a circle from a point T outside it, then TA = TB. Reason to write: “tangents from an external point are equal in length”. Why? Triangles OAT and OBT are right-angled (Fact 1) and share the hypotenuse OT. By Pythagoras, TA² = OT² − OA² and TB² = OT² − OB². Since OA = OB (radii) the two lengths are equal.

The shape OATB is a kite: two pairs of equal sides (OA = OB and TA = TB) and two right angles at A and B.

The kite. The quadrilateral OATB is a kite. Its two right angles at A and B mean the other two angles add up to 180°: angle AOB + angle ATB = 180°. The line OT is the line of symmetry. It bisects angle ATB and angle AOB, and it is perpendicular to the chord AB. Triangle TAB is isosceles (because TA = TB), and triangle OAB is isosceles (because OA = OB).

Key ideas

A tangent is perpendicular to the radius at the point of contact. Reason: “the angle between a tangent and a radius is 90°”.
Tangents from an external point are equal in length. Reason: “tangents from an external point are equal in length”.
The right angle makes Pythagoras' theorem available: (tangent length)² + (radius)² = (distance from the centre)².
Quadrilateral OATB is a kite: angle AOB + angle ATB = 180°.
Triangles TAB and OAB are both isosceles.

Watch out

The right angle is between the tangent and the radius, at the point of contact. It is not at the external point T, and it is not between the two tangents.

Worked example 26.6.1

TA is a tangent to the circle with centre O, touching the circle at A. Angle ATO = 30°. Work out the size of angle AOT. Give a reason for each step.

Key words. A tangent is a straight line that touches a circle at exactly one point, the point of contact.

Method. A tangent is at right angles to the radius at the point where it touches the circle. That makes triangle OAT right-angled.

Draw the triangle. Join O to A and to T. The radius OA, the tangent TA and the line OT make triangle OAT.
triangle OAT
Find the angle at the point of contact. A tangent touches the circle at one point and is perpendicular to the radius there.
Angle OAT = 90° (a tangent is perpendicular to the radius at the point of contact)
Use the angle sum of triangle OAT. The three angles add up to 180°. Subtract the two we know.
Angle AOT = 180° − 90° − 30° = 60° (the angles in a triangle add up to 180°)

Check. 60° + 30° + 90° = 180°. ✓

Watch out. The right angle is at A, the point of contact, between the radius and the tangent.

Worked example 26.6.2

TA and TB are tangents to a circle with centre O. Angle ATB = 64°. Work out the size of angle AOB. Give a reason for each step.

Method. The two radii and the two tangents make a quadrilateral OATB with two right angles. Use the angle sum 360°.

Draw the radii OA and OB. They go to the two points of contact and make a quadrilateral OATB with the two tangents.
quadrilateral OATB
Mark the right angles. Each tangent is perpendicular to the radius drawn to its point of contact.
Angle OAT = 90° and angle OBT = 90° (a tangent is perpendicular to the radius at the point of contact)
Use the quadrilateral OATB. Its four angles add up to 360°. Subtract the three we know.
Angle AOB = 360° − 90° − 90° − 64° = 116° (the angles in a quadrilateral add up to 360°)

Check. 360° = 360°. ✓ A shortcut for later: angle AOB + angle ATB = 180°.

Watch out. The quadrilateral OATB is a kite. Its two right angles are at A and B, opposite each other.

Worked example 26.6.3

TA and TB are tangents to a circle with centre O. Angle ATB = 50°. Work out the size of angle TAB. Give a reason for each step.

Method. Two tangents from one point are equal in length, so triangle TAB is isosceles.

Show triangle TAB is isosceles. Tangents drawn from the same point to a circle are equal in length.
TA = TB (tangents from an external point are equal in length)
Write the base angles. The base angles of an isosceles triangle are equal: the base angles are at A and B.
Angle TAB = angle TBA
Use the angle sum. Subtract the angle at T and share the rest equally.
Angle TAB = (180° − 50°) ÷ 2 = 130° ÷ 2 = 65° (the angles in a triangle add up to 180°)

Check. 65° + 65° + 50° = 180°. ✓

Watch out. Say why the triangle is isosceles: the tangents from T are equal. Do not just say "it looks isosceles".

Worked example 26.6.4

TA is a tangent to a circle with centre O, touching it at A. The radius of the circle is 5 cm and OT = 13 cm. Work out the length of TA.

Method. The radius meets the tangent at a right angle, so triangle OAT is right-angled and we can use Pythagoras' theorem.

Find the right angle. A tangent is perpendicular to the radius at the point of contact.
Angle OAT = 90° (a tangent is perpendicular to the radius at the point of contact)
Write Pythagoras for triangle OAT. The longest side is OT, opposite the right angle.
OT² = OA² + TA²
13² = 5² + TA²
Rearrange. Take the known square away from both sides.
TA² = 169 − 25 = 144
Take the square root. A length is positive.
TA = √144 = 12 cm

Check. 5² + 12² = 169, which is 13². ✓

Watch out. OT is the hypotenuse, so it is the one you square and then subtract from, not add to.

Worked example 26.6.5

TA and TB are tangents to a circle with centre O. TA = (3x + 2) cm and TB = (5x − 6) cm.

Work out the value of x.
Work out the length of TA.

Method. Tangents from an external point are equal in length, so the two expressions are equal. Solve the equation, then substitute.

Use the tangent fact. Tangents drawn from the same point to a circle are equal in length.
TA = TB (tangents from an external point are equal in length)
Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.
5x − 6 = 3x + 2 (written the other way round so that the x terms stay positive)
Subtract 3x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.
5x − 6 − 3x = 3x + 2 − 3x
2x − 6 = 2
Add 6 to both sides. We want the number terms on the other side.
2x = 2 + 6
2x = 8
Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.
x = 8 ÷ 2 = 4
Work out the length. Put x back into the expression for TA.
TA = 3 × 4 + 2 = 14 cm

Check. TB = 5 × 4 − 6 = 14 cm, the same as TA. ✓

Watch out. Check the final length is positive. A negative length means a mistake.

Worked example 26.6.6

TA and TB are tangents to a circle with centre O. Prove that TA = TB.

Method. Use the right angles at the points of contact to apply Pythagoras' theorem to both triangles.

Draw the line OT. It is the hypotenuse of both triangles OAT and OBT.
triangles OAT and OBT share the side OT
Find the right angles. A tangent is perpendicular to the radius at the point of contact.
angle OAT = angle OBT = 90° (the angle between a tangent and a radius is 90°)
Write Pythagoras for each triangle. In a right-angled triangle, the hypotenuse squared is the sum of the other two sides squared.
OT² = OA² + TA² and OT² = OB² + TB²
Use the equal radii. OA and OB are radii of the same circle.
OA = OB, so OA² = OB²
Compare the two equations. The left sides are the same, and the first term on each right side is the same, so the second terms must be equal.
TA² = OT² − OA² = OT² − OB² = TB²
Take the square root. Lengths are positive.
TA = TB. ✓

Check. Test: radius 5, OT = 13. Then TA = √(169 − 25) = 12 and TB = 12. ✓

Watch out. A second proof uses congruent triangles: the triangles have a right angle, the same hypotenuse and one equal side (RHS), so they are congruent.

26.7

The alternate segment theorem

Edexcel 1MA1: G10
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The alternate segment theoremComing soon

Draw a circle with a tangent PQ touching it at T, and draw a chord TA starting at the point of contact. The chord makes an angle with the tangent. The chord also cuts the circle into two segments. The segment on the other side of the chord from that angle is called the alternate segment. Pick any point B on the circle in that segment and join it to T and A. The theorem says that angle TBA is equal to the angle between the tangent and the chord.

Angle QTA (between the tangent and the chord TA) = angle TBA (in the alternate segment). Both are marked x.

Reason to write: “the angle between a tangent and a chord is equal to the angle in the alternate segment”. A good way to use it: (1) find the chord that starts at the point of contact; (2) decide which side of that chord the tangent angle is on; (3) the equal angle is made at a point on the other side. There are two tangent-chord angles at T (one on each side of the chord), and each equals a different angle in the opposite segment.

Why is it true? Draw the diameter TD through the centre and join D to A. Let angle QTA = x. The tangent is perpendicular to the diameter at T, so angle QTD = 90° and angle DTA = 90° − x. Angle TAD is an angle in a semicircle, so it is 90°. In triangle TAD the third angle is 180° − 90° − (90° − x) = x, so angle TDA = x. Finally, angle TBA and angle TDA are in the same segment, so angle TBA = x as well. The full proof is worked example 26.7.5.

The proof: TD is a diameter. The right angles come from the tangent theorem (at T) and the semicircle theorem (at A).

Key ideas

The angle between a tangent and a chord is equal to the angle in the alternate segment.
Reason: “the angle between a tangent and a chord is equal to the angle in the alternate segment”.
The alternate segment is on the opposite side of the chord from the tangent angle.
Each of the two angles at the point of contact has its own partner: tangent angle QTA goes with TBA; tangent angle PTB goes with TAB.
The angle between the tangent and a chord is half of the angle at the centre on the arc that angle faces.

Watch out

The equal angle is at the other side of the chord. Do not match the tangent angle with the angle on the same side of the chord. A good check: the two equal angles should not be next to each other.

Worked example 26.7.1

PQ is a tangent to the circle at T. A and B are points on the circle. Angle QTA = 40°. Work out the size of angle TBA. Give a reason for your answer.

Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.

Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).

Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.
Angle between tangent TQ and chord TA = 40°
Find the alternate segment. The alternate segment is the part of the circle on the other side of the chord TA from the tangent angle. That is where B is, so angle TBA is in the alternate segment.
angle TBA is in the alternate segment of the chord TA
Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Angle TBA = angle QTA = 40° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. Angle at the centre on arc TA would be 2 × 40° = 80°. Both angles are half of it. ✓

Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.

Worked example 26.7.2

PQ is a tangent to the circle at T. A and B are points on the circle. Angle QTA = 35° and angle PTB = 75°. Give a reason for each answer.

Work out angle TBA
Work out angle TAB
Work out angle ATB

Part (a)

Method. Use the alternate segment theorem: the angle between the tangent and a chord equals the angle in the alternate segment.

Look at the tangent angle on the chord TA. Angle QTA is between the tangent and chord TA. The alternate segment is on the other side of TA, and it contains B.
angle QTA = 35°, and the matching angle is angle TBA
Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Angle TBA = angle QTA = 35° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. The matching angle at the centre on arc TA would be 70°. ✓

Watch out. The alternate segment is the one across the chord from the tangent angle.

Part (b)

Method. Use the alternate segment theorem again, this time with the chord TB.

Look at the tangent angle on the chord TB. Angle PTB is between the tangent and chord TB. The alternate segment is on the other side of TB, and it contains A.
angle PTB = 75°, and the matching angle is angle TAB
Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.
Angle TAB = angle PTB = 75° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. The matching angle at the centre on arc TB would be 150°. ✓

Watch out. Use the angle on the opposite side of chord TB from PTB, which is at A.

Part (c)

Method. The three angles of triangle ATB add up to 180°.

Write the two angles we know. We found them in parts (a) and (b).
Angle TBA = 35° and angle TAB = 75°
Subtract from 180°. The angles in a triangle add up to 180°.
Angle ATB = 180° − 35° − 75° = 70°

Check. Also angle PTB + angle ATB + angle QTA = 75 + 70 + 35 = 180 (angles on a straight line). ✓

Watch out. You can also find this angle from the straight line PQ.

Worked example 26.7.3

PQ is a tangent at T to a circle with centre O. Angle QTA = 38°. Work out the size of angle TOA. Give a reason for each step.

Method. First use the alternate segment theorem to get an angle at the circumference, then double it for the angle at the centre.

Find the angle in the alternate segment. Angle QTA is between the tangent and the chord TA.
Angle TBA = 38° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Name the arc. Angle TOA is at the centre and angle TBA is at the circumference. Both stand on the arc TA.
arc TA
Use the angle at the centre. The angle at the centre is twice the angle at the circumference.
Angle TOA = 2 × 38° = 76° (the angle at the centre is twice the angle at the circumference)

Check. Second route: angle OTQ = 90° (tangent and radius), so angle OTA = 90° − 38° = 52°. Triangle OTA is isosceles, so angle OAT = 52° and angle TOA = 180° − 2 × 52° = 76°. ✓

Watch out. You can also find this answer without the alternate segment theorem, using the right angle at T (see the check).

Worked example 26.7.4

PQ is a tangent to the circle at T. Angle QTA = (3x + 4)° and angle TBA = (x + 28)°. The two angles are marked on the diagram.

Work out the value of x.
Work out the size of angle QTA.

Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.

Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.
(3x + 4) = (x + 28) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.
3x + 4 = x + 28
Subtract x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.
3x + 4 − x = x + 28 − x
2x + 4 = 28
Subtract 4 from both sides. We want the number terms on the other side.
2x = 28 − 4
2x = 24
Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.
x = 24 ÷ 2 = 12
Work out the angle. Put x into the expression for angle QTA.
Angle QTA = 3 × 12 + 4 = 40°

Check. Other angle: 1 × 12 + 28 = 40°, equal to the first. ✓

Watch out. These angles are equal, they do not add up to 90° or 180°.

Worked example 26.7.5

Prove the alternate segment theorem. PQ is a tangent to a circle at T. A and B are points on the circle. Show that angle QTA = angle TBA.

Method. Draw the diameter TD and join DA. Find angle TDA in two ways: using the angle sum and using the same-segment theorem.

Name the tangent angle. Use a letter so the proof works for any size of angle.
Let angle QTA = x
Use the tangent and the diameter. TD is a diameter, so it lies along the radius at T. A tangent is perpendicular to the radius at the point of contact.
angle QTD = 90°, so angle DTA = 90° − x
Use the semicircle theorem. TD is a diameter and A is on the circle.
angle TAD = 90° (angle in a semicircle is 90°)
Use the angle sum in triangle TAD. The angles add up to 180°.
angle TDA = 180° − 90° − (90° − x) = x
Move to the point B. Angles TDA and TBA are made from the same chord TA and are on the same side of it, so they are in the same segment.
angle TBA = angle TDA = x (angles in the same segment are equal)
Conclude.
angle TBA = x = angle QTA. ✓

Check. Test with x = 40°: angle DTA = 50°, angle TAD = 90°, angle TDA = 40°, and 50 + 90 + 40 = 180. ✓

Watch out. In this proof the point D is in the alternate segment. If B is elsewhere in that segment, the same-segment theorem carries the result to B.

26.8

Chords

Edexcel 1MA1: G10
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: ChordsComing soon

A chord joins two points on a circle. There is a neat fact about the line from the centre that meets the chord at a right angle: the perpendicular from the centre to a chord bisects the chord. Bisects means cuts exactly in half. So if M is the point where the perpendicular from the centre O meets the chord AB, then AM = MB.

If OM is perpendicular to the chord AB, then M is the midpoint of AB. The tick marks show AM = MB.

Reason to write: “the perpendicular from the centre to a chord bisects the chord”. The opposite is also true: a line from the centre to the midpoint of a chord is perpendicular to the chord. Why? Join O to A and B. These are radii, so OA = OB and triangle OAB is isosceles. The line from the apex O that meets the base at 90° splits the triangle into two identical right-angled triangles (they share OM, have equal hypotenuses OA = OB, and a right angle at M). So AM = MB.

The proof: OA = OB (radii, double ticks) and the right angle at M make two congruent triangles OAM and OBM.

Using it with Pythagoras. The right-angled triangle OMA has the radius as its hypotenuse, half the chord as one short side and the distance from the centre to the chord as the other short side: r² = d² + (half the chord)². Given any two of the three lengths, you can find the third. Remember to halve the chord before you use Pythagoras, and to double at the end if you want the whole chord.

Using it with trigonometry. The perpendicular also halves the angle at the centre: triangle OAB is isosceles, so OM bisects angle AOB. In triangle OMA you can use sin, cos or tan with the half chord, the radius and the half angle (Chapter 20). The calculator exercise at the end of this section practises this.

Key ideas

The perpendicular from the centre to a chord bisects the chord, and a line from the centre to the midpoint of a chord is perpendicular to it.
Reason: “the perpendicular from the centre to a chord bisects the chord”.
Right-angled triangle: hypotenuse = radius, short sides = half the chord and the distance from the centre.
r² = d² + (half chord)².
The depth of a segment (the height of the cut-off part) = radius − d, when the segment is the smaller part.

Watch out

Use half the chord in Pythagoras. The most common mistake is to use the full chord as a side of the right-angled triangle.

Worked example 26.8.1

AB is a chord of a circle with centre O and radius 13 cm. AB = 24 cm. M is the point where the perpendicular from O meets the chord. Work out the distance OM.

Key words. To bisect means to cut exactly in half. A perpendicular meets another line at 90°.

Method. The perpendicular from the centre cuts the chord in half, which makes a right-angled triangle with the radius as hypotenuse.

Find the length AM. The perpendicular from the centre to a chord bisects (cuts in half) the chord.
AM = 24 ÷ 2 = 12 cm (the perpendicular from the centre to a chord bisects the chord)
Write Pythagoras for triangle OMA. The angle at M is 90°. OA is a radius, and it is the longest side (the hypotenuse).
OA² = OM² + AM²
13² = OM² + 12²
Rearrange. Take the known square from both sides.
OM² = 169 − 144 = 25
Square root.
OM = √25 = 5 cm

Check. 5² + 12² = 25 + 144 = 169 = 13². ✓

Watch out. Use half the chord in the triangle, not the whole chord.

Worked example 26.8.2

AB is a chord of a circle with centre O and radius 10 cm. The perpendicular distance from O to the chord is 6 cm. Work out the length of the chord AB.

Method. The perpendicular from the centre bisects the chord. Find half the chord with Pythagoras, then double.

Draw the triangle. The perpendicular from the centre meets the chord at M and makes a right-angled triangle OMA with OA as the hypotenuse.
OA = 10 cm, OM = 6 cm
Use Pythagoras to find AM. Subtract the square of the known short side from the square of the hypotenuse.
AM² = 10² − 6² = 100 − 36 = 64
AM = √64 = 8 cm
Double it to get the whole chord. The perpendicular from the centre bisects the chord, so AM is only half of it.
AB = 2 × 8 = 16 cm (the perpendicular from the centre to a chord bisects the chord)

Check. 6² + 8² = 100, so OA = 10. ✓

Watch out. The triangle gives you half the chord. Remember to double.

Worked example 26.8.3

AB is a chord of length 30 cm in a circle with centre O. The perpendicular distance from O to the chord is 8 cm. Work out the radius of the circle.

Method. Half the chord and the distance from the centre are the two short sides of a right-angled triangle. The radius is its hypotenuse.

Find half the chord. The perpendicular from the centre to a chord bisects the chord.
AM = 30 ÷ 2 = 15 cm (the perpendicular from the centre to a chord bisects the chord)
Use Pythagoras in triangle OMA. The angle at M is 90°, and the radius OA is the hypotenuse.
OA² = 8² + 15² = 64 + 225 = 289
Square root. The radius is positive.
OA = √289 = 17 cm

Check. The distance from the centre to B is the same: √(8² + 15²) = 17. ✓

Watch out. Add the squares to get the hypotenuse.

Worked example 26.8.4

A circular pipe has radius 13 cm. Water fills the pipe to a level so that the water surface is a chord of length 24 cm. Work out the depth of the water at its deepest point (the vertical distance from the chord to the lowest part of the circle).

Method. Use the perpendicular from the centre to the chord, then take the distance from the radius.

Draw the picture. The surface of the water is a chord. The lowest point of the water is at the bottom of the circle, directly below the centre.
Chord = 24 cm, radius = 13 cm
Find the distance from the centre to the chord. The perpendicular from the centre bisects the chord, which gives a right-angled triangle with half the chord and the radius.
Half the chord = 12 cm
Distance² = 13² − 12² = 169 − 144 = 25
Distance = 5 cm
Work out the depth. The depth is the radius take away the distance from the centre to the chord.
Depth = 13 − 5 = 8 cm

Check. Water is 8 cm deep and the centre is 5 cm above the surface: 8 + 5 = 13 cm, the radius. ✓

Watch out. The depth is not the distance from the centre to the chord. It is the radius minus that distance.

Worked example 26.8.5

AB is a chord of a circle with centre O. M is the midpoint of AB. Prove that OM is perpendicular to AB.

Method. Show triangles OAM and OBM are congruent, then the angles at M are equal and must each be 90°.

Join OA and OB. These are radii, so they are equal.
OA = OB (radii)
Use the midpoint. The question says M is the midpoint of AB.
AM = MB
Shared side. OM is in both triangles.
OM is common
Congruent triangles. All three pairs of sides are equal (SSS).
Triangles OAM and OBM are congruent (SSS).
Match the angles at M. Congruent triangles have equal matching angles.
angle OMA = angle OMB
Use the straight line. A, M and B are in a straight line, so these two equal angles add up to 180°.
each is 180° ÷ 2 = 90°. So OM is perpendicular to AB. ✓

Check. Test: radius 13, AM = MB = 12, OM = 5. Then 5² + 12² = 169 = 13², so the angle at M is 90°. ✓

Watch out. The proof shows the second half of the theorem: from the midpoint to the right angle.

26.9

Problems, algebra and proof

Edexcel 1MA1: G9, G10
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Problems, algebra and proofComing soon

Exam questions rarely use one theorem on its own. Most ask you to chain two or three facts, add algebra, or prove a result. This section gives you a method for them.

A method for circle theorem problems

1. Mark the diagram. Write every angle you are told on it. Mark equal radii with ticks and mark the centre.
2. Look for the clues. Centre or radii (isosceles triangle, centre theorem). Diameter (90°). Tangent (90° with the radius, equal tangents, alternate segment). Four points on a circle (cyclic quadrilateral, same segment).
3. Find one new angle at a time. Write it with its reason in the standard wording, then use it for the next step.
4. If there is algebra, write the angle fact as an equation (equal angles: expression = expression; adding to 180°: sum = 180), solve, then substitute back.
5. Check. All angles in a triangle sum to 180°; try a second route if you can.

Show that and prove. In a “show that” or “prove” question the answer is given, so every step and its reason must be written down. A good layout is a list of statements, each with a reason: “angle OAB = 90° − x (base angle of an isosceles triangle …)”. Use letters instead of numbers so the proof is true for every case. End with a clear statement of what you have shown. Marks go for the method, the reasons and the conclusion. Do not use the result you are trying to prove.

Exam tip

Always write the reason next to the angle, in the standard words from Section 26.1. A correct angle with no reason scores fewer marks than a correct angle with a reason.

Watch out

Do not use “it looks like” or measure the diagram. Diagrams in the exam are not accurate. Only use what you have been told or what you can prove from the theorems.

Worked example 26.9.1

TU is a tangent to a circle with centre O, touching the circle at A. B is a point on the circle. Angle UAB = 38°. Give a reason for each step.

Work out angle OAB
Work out angle AOB

Part (a)

Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.

Find the right angle. Radius OA meets the tangent TU at the point of contact A.
Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
Subtract. Angle OAU is made of angle OAB and angle BAU.
Angle OAB = 90° − 38° = 52°

Check. 52° + 38° = 90°. ✓

Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.

Part (b)

Method. Triangle OAB has two equal sides (both radii), so it is isosceles.

Write the base angles. OA and OB are radii, so the base angles are equal.
Angle OBA = angle OAB = 52° (the base angles of an isosceles triangle are equal)
Use the angle sum. Subtract the two base angles from 180°.
Angle AOB = 180° − 52° − 52° = 76° (the angles in a triangle add up to 180°)

Check. Alternate segment theorem: the angle in the alternate segment is 38°, and the angle at the centre is 2 × 38° = 76°. ✓

Watch out. The apex angle of the isosceles triangle is at the centre O.

Worked example 26.9.2

AB is a diameter of a circle with centre O. TU is a tangent to the circle at A. C is a point on the circle. Angle UAC = 34°. Give a reason for each answer.

Work out angle CAB.
Work out angle ACB.
Work out angle ABC.

Part (a)

Method. A tangent is perpendicular to the radius at the point of contact. AB is a diameter, so it contains the radius OA.

Find angle UAB. The diameter AB is along the radius OA, and the tangent TU touches the circle at A.
Angle UAB = 90° (a tangent is perpendicular to the radius at the point of contact)
Subtract. Angle UAB is made of angle UAC and angle CAB side by side.
Angle CAB = 90° − 34° = 56°

Check. 34° + 56° = 90°. ✓

Watch out. The right angle is between the tangent and the diameter, so it contains both angles.

Part (b)

Method. AB is a diameter, so angle ACB is an angle in a semicircle.

Spot the diameter. The two arms CA and CB go to the two ends of the diameter AB.
Angle ACB = 90° (angle in a semicircle is 90°)
Check the triangle. Triangle ABC now has a right angle at C.
angle ACB = 90°

Check. Angle CAB + angle ACB = 56° + 90° = 146°, which is less than 180°, so there is room for angle ABC. ✓

Watch out. The right angle is at C, on the circle.

Part (c)

Method. Use the angle sum in triangle ABC, then check with the alternate segment theorem.

Use the angles we know. Angle CAB is from part (a) and angle ACB is from part (b).
Angle ABC = 180° − 56° − 90° = 34° (the angles in a triangle add up to 180°)
Check with the alternate segment theorem. Angle UAC is between the tangent and the chord AC. The angle in the alternate segment on the other side of AC is angle ABC.
Angle ABC = angle UAC = 34° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. Both methods give 34°. ✓

Watch out. When two different routes give the same answer, you can be confident it is right.

Worked example 26.9.3

A, B, C and D are points on a circle. PQ is a tangent to the circle at A. Angle QAB = 50° and angle PAD = 45°. Give a reason for each answer.

Work out angle ACB.
Work out angle BAD.
Work out angle BCD.

Part (a)

Method. Use the alternate segment theorem with the tangent PQ and the chord AB.

Find the angle between the tangent and the chord. Angle QAB is between the tangent AQ and the chord AB.
angle QAB = 50°
Find the alternate segment. The alternate segment is on the other side of the chord AB, where C and D are. So the angle at C matches.
Angle ACB = angle QAB = 50° (the angle between a tangent and a chord is equal to the angle in the alternate segment)

Check. Angle ADB is also 50° (same segment), so both corners agree. ✓

Watch out. The alternate segment is the one on the opposite side of the chord from the tangent angle.

Part (b)

Method. P, A and Q are on a straight line (the tangent), so the three angles at A add up to 180°.

Write the three angles at A. They are angle PAD, angle DAB and angle BAQ on the line PQ.
45° + angle BAD + 50° = 180°
Subtract.
Angle BAD = 180° − 45° − 50° = 85° (angles on a straight line add up to 180°)

Check. 45 + 85 + 50 = 180. ✓

Watch out. The angle between the two chords is the middle angle on the straight line.

Part (c)

Method. ABCD is a cyclic quadrilateral and A is opposite C.

Use the answer to part (b). Angle BAD is the interior angle at A.
Angle BAD = 85°
Use the cyclic quadrilateral rule. Opposite angles of a cyclic quadrilateral add up to 180°.
Angle BCD = 180° − 85° = 95° (opposite angles of a cyclic quadrilateral add up to 180°)

Check. 85° + 95° = 180°. ✓

Watch out. Angle BCD is opposite angle BAD, not next to it.

Worked example 26.9.4

ABC is a triangle with AB = AC. Its corners are on a circle with centre O. Angle BAC = 40°.

Work out angle BOC.
Work out angle OBC.
Work out angle ABC.
Work out angle ABO.

Part (a)

Method. BAC is at the circumference and BOC is at the centre. Both stand on the arc BC.

Name the arc. Both angles are made from the points B and C.
arc BC
Double it. The angle at the centre is twice the angle at the circumference.
Angle BOC = 2 × 40° = 80° (the angle at the centre is twice the angle at the circumference)

Check. Half of 80° is 40°. ✓

Watch out. Use the circumference angle you are given and double it.

Part (b)

Method. OB and OC are radii, so triangle OBC is isosceles.

Spot the isosceles triangle. Radii are equal, so the base angles are equal.
angle OBC = angle OCB (the base angles of an isosceles triangle are equal)
Use the angle sum. Subtract the angle at the centre from 180° and halve.
Angle OBC = (180° − 80°) ÷ 2 = 50° (the angles in a triangle add up to 180°)

Check. 50° + 50° + 80° = 180°. ✓

Watch out. The apex of this triangle is at the centre O.

Part (c)

Method. Triangle ABC is isosceles because AB = AC.

Write the base angles. The equal sides are AB and AC, so the base angles at B and C are equal.
angle ABC = angle ACB (the base angles of an isosceles triangle are equal)
Use the angle sum. The angle at A is 40°.
Angle ABC = (180° − 40°) ÷ 2 = 70° (the angles in a triangle add up to 180°)

Check. 70° + 70° + 40° = 180°. ✓

Watch out. Here the apex is at A, on the circle, not at the centre.

Part (d)

Method. Angle ABC is made of angle ABO and angle OBC side by side.

Write the sum. The radius BO lies inside angle ABC.
angle ABO + angle OBC = angle ABC
Subtract. Use angle ABC from part (c) and angle OBC from part (b).
Angle ABO = 70° − 50° = 20°

Check. Triangle OAB is isosceles with OA = OB, so angle OAB = angle OBA = 20°, and AO bisects angle BAC = 40°. ✓

Watch out. Subtract the smaller part from the whole angle.

Worked example 26.9.5

TA and TB are tangents to a circle with centre O. Prove that angle ATB = 180° − angle AOB. Then show that OT bisects angle ATB.

Method. Use the right angles at the points of contact in the quadrilateral OATB, then the congruent triangles.

Find the angles at A and B. A tangent is perpendicular to the radius at the point of contact.
angle OAT = angle OBT = 90° (the angle between a tangent and a radius is 90°)
Use the angle sum of the quadrilateral. The angles in a quadrilateral add up to 360°.
angle AOB + angle ATB + 90° + 90° = 360°
Rearrange. Subtract 180° from both sides.
angle AOB + angle ATB = 180°, so angle ATB = 180° − angle AOB. ✓
Now show OT bisects angle ATB. Triangles OAT and OBT have a right angle, a common hypotenuse OT and OA = OB (radii).
They are congruent (RHS).
Match the angles at T. Congruent triangles have equal matching angles.
angle ATO = angle BTO, so OT bisects angle ATB. ✓

Check. Test: if angle ATB = 64°, then angle AOB = 116° and each half at T is 32°. 64 + 116 = 180. ✓

Watch out. Use the standard reason after each line, for example “angles in a quadrilateral add up to 360°”.

Exercise 26K Reasoning and proof

Non-calculator

RName the circle theorem that is being used in each statement.

(a)AB is a diameter and C is on the circle, so angle ACB = 90°.

(b)Angle AOB = 120° at the centre and angle ACB = 60° at the circumference.

(c)Angle BAC = angle BDC, where A and D are on the same side of the chord BC.

(d)ABCD is a cyclic quadrilateral so angle ABC + angle ADC = 180°.

(e)TA is a tangent and OA is a radius, so angle OAT = 90°.

(f)TA and TB are tangents from T, so TA = TB.

(g)PQ is a tangent at T and angle QTA = angle TBA.

(h)M is the point where the perpendicular from O meets the chord AB, so AM = MB.

Done

RComplete each reason using words from the box: radii, diameter, tangent, cyclic quadrilateral, same segment, semicircle.

(a)Triangle OAB is isosceles because OA and OB are ______.

(b)Angle ABC = 90° because AC is a ______.

(c)Angle OAT = 90° because TA is a ______.

(d)Angle ABD = angle ACD because they are angles in the ______.

Done

RABCD is a cyclic quadrilateral with AB = AD. The diagonals AC and BD are drawn. Prove that CA bisects angle BCD.

Done

RTA and TB are tangents to a circle and C is a point on the major arc AB. Prove that angle ACB = 90° − ½ angle ATB.

Done

RABCD is a cyclic quadrilateral with AB parallel to DC. Prove that AD = BC.

Done

RIn the diagram, O is the centre of the circle, and AB is a chord. A tangent at A makes an angle x with the chord AB. Show that angle AOB = 2x in two different ways.

Done

Chapter summary

Skill	Method
Circle words	Radius (centre to circle), diameter (chord through centre), chord (joins two points on the circle), arc, sector (two radii), segment (a chord), tangent (touches at one point)
Radii triangle	Two radii are equal, so triangle OAB is isosceles: base angles equal, angles add up to 180°
Angle in a semicircle	If AB is a diameter and C is on the circle then angle ACB = 90°
Angle at the centre	Centre angle = 2 × circumference angle on the same arc. Minor-arc point: use the reflex angle
Same segment	Angles made by the same chord with corners on the same side of it are equal
Cyclic quadrilateral	Opposite angles add up to 180°; exterior angle = interior opposite angle
Tangent and radius	The angle between a tangent and the radius at the point of contact is 90°; use Pythagoras for lengths
Two tangents	Tangents from an external point are equal; OATB is a kite; angle AOB + angle ATB = 180°
Alternate segment	Angle between a tangent and a chord = angle in the alternate segment (the opposite side of the chord)
Chords	The perpendicular from the centre bisects the chord; r² = d² + (half chord)²
Algebra with angles	Equal angles: expression = expression. Angles adding to 180° (or 90°): sum of expressions = 180. Solve, then substitute back
Proof	Use letters, give a reason for each line, finish with a conclusion

Can you…?

I can name and describe a chord, tangent, arc, sector and segment.
I can use the angle in a semicircle and explain why it is 90°.
I can use and prove that the angle at the centre is twice the angle at the circumference, including with a reflex angle.
I can use angles in the same segment and find the equal pairs in a diagram with crossing chords.
I can use opposite angles and exterior angles of a cyclic quadrilateral, and prove the theorem.
I can use a tangent being perpendicular to the radius, and tangents from a point being equal, with angles and Pythagoras.
I can use and prove the alternate segment theorem.
I can use that the perpendicular from the centre bisects a chord to find lengths and angles.
I can combine theorems with isosceles triangles and algebra, giving a reason for every step in standard wording.
I can write a clear proof or “show that” argument with letters, reasons and a conclusion.

Key terms

Chord: A straight line that joins two points on a circle.
Diameter: A chord that passes through the centre. It is twice the radius.
Radius: A straight line from the centre of a circle to a point on the circle.
Arc: A part of the circumference. The shorter is the minor arc, the longer is the major arc.
Sector: The region between two radii and an arc.
Segment: The region between a chord and an arc.
Tangent: A straight line that touches a circle at exactly one point (the point of contact).
Circumference: The distance all the way round a circle, or the circle itself.
Cyclic quadrilateral: A four-sided shape with all four corners on a circle.
Alternate segment: The segment on the opposite side of a chord from a given angle.
Bisect: To cut exactly in half.
Subtend / stand on: An angle stands on an arc if its arms end at the two ends of that arc.
Reflex angle: An angle bigger than 180° and smaller than 360°.
Kite: A quadrilateral with two pairs of equal adjacent sides. OATB is a kite when TA and TB are tangents.
Converse: The statement the other way round: if the angle is 90°, then the line is a diameter.

Total 59 marks. Allow about 59 minutes.

← Chapter 25: Functions and advanced graphs

GCSE Maths Circle Theorems (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Circle words and isosceles triangles

Words to know

The reasons you must be able to write

Watch out

Exercise 26A Circle words and isosceles triangles

The angle in a semicircle

Key ideas

Watch out

Exercise 26B The angle in a semicircle

The angle at the centre

Key ideas

Watch out

Exercise 26C The angle at the centre

Angles in the same segment

Key ideas

Watch out

Exercise 26D Angles in the same segment

Cyclic quadrilaterals

Key ideas

Watch out

Exercise 26E Cyclic quadrilaterals

Tangents and radii

Key ideas

Watch out

Exercise 26F Tangents and radii

The alternate segment theorem

Key ideas

Watch out

Exercise 26G The alternate segment theorem

Chords

Key ideas

Watch out

Exercise 26H Chords (no calculator)

Exercise 26I Chords, tangents and trigonometry (calculator allowed)

Problems, algebra and proof

A method for circle theorem problems

Exam tip

Watch out

Exercise 26J Multi-step circle theorem problems

Exercise 26K Reasoning and proof

Chapter summary

Can you…?

Key terms

Exercise 26Z Check-up (non-calculator)

Exercise 26X Exam-style questions