Combining circle theorems — exam problems — Mathematics

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This leads to

Using vectors to prove geometric results · coming soon
Arc length and sector area · coming soon
Tangents and normals to a circle · coming soon

You know every theorem. So why is this hard?

Look at the figure below. Four points A, B, C and D lie on a circle. A line XAY touches the circle at A. You are given two angles at A, 65° and 48°. Find the angle BCD.

An exam-style problem. XAY is the tangent at A. Find ∠BCD. No theorem gives the answer in one step.

You have met every theorem this problem needs. Yet many students look at a diagram like this and do not know where to start. That is not a sign that they have forgotten the theorems. An exam diagram hides three or four theorems, and you must find them, choose an order, and connect them into a chain.

So the difficulty does not live in any single theorem. It lives in the route. This lecture is about finding routes. By the end, you will solve the problem above, and you will also see why its answer, 113°, is exactly 65° + 48°.

The toolkit on one page

Each circle theorem is switched on by something you can see in the diagram. The table pairs what you see with what it gives you, and with the reason you write in an exam answer.

What you see	What it gives you	The reason you write
An angle at the centre O and an angle at the circumference, standing on the same arc	The angle at O is twice the other	Angle at the centre is twice the angle at the circumference
Two angles at the circumference standing on the same arc, from the same side	The two angles are equal	Angles in the same segment are equal
A diameter, and a point on the circle joined to both its ends	A right angle at that point	Angle in a semicircle is 90°
Four corners of a quadrilateral, all on the circle	Opposite angles add up to 180°	Opposite angles of a cyclic quadrilateral
A tangent, and a radius to its point of contact	A right angle at the point of contact	Tangent is perpendicular to the radius
Two tangents from one outside point	Equal lengths, so an isosceles triangle	Tangents from a point are equal
A tangent and a chord that meet at the point of contact	The angle between them equals the angle in the alternate segment	Alternate segment theorem

Circle theorems never work alone. They pass their answers to older tools: angles in a triangle, on a straight line and in a quadrilateral, and parallel lines. One older tool deserves special attention. Two radii are always equal, so any triangle with a corner at the centre and two corners on the circle is isosceles, even when nobody says so.

Now practise the first skill on its own: seeing the signpost. Each card below describes something in a diagram. Sort it into the theorem it switches on.

Sort — which theorem does this switch on?

Read each card. Choose the box for the theorem it switches on. Two cards describe a trap: put them in the last box.

This table pairs each feature of a diagram with the theorem it switches on, and what that theorem gives you.

In the diagram	Theorem
AB is labelled as a diameter, and C is on the circle.	Angle in a semicircle: ∠ACB = 90°.
A, B, C and D all lie on the circle.	Cyclic quadrilateral: opposite angles add up to 180°.
PT touches the circle at T, and the radius OT is drawn.	Tangent and radius: ∠OTP = 90°.
PA and PB touch the circle at A and B.	Two tangents: PA = PB, so triangle PAB is isosceles.
The tangent at T and the chord TA meet at T.	Alternate segment theorem.
∠AOB at the centre and ∠ACB at C stand on the same arc AB.	Angle at the centre: ∠AOB = 2∠ACB.
∠ADB and ∠ACB stand on AB, with C and D on the same side.	Angles in the same segment are equal.
Quadrilateral OACB has a corner at the centre O.	Trap: O is not on the circle, so OACB is not a cyclic quadrilateral of this circle.
A chord looks as if it passes through the centre, but the question does not say so.	Trap: you may not assume it is a diameter.

The two trap cards matter most. A theorem can be used only when its trigger is truly in the diagram.

Stop and think

Go back to the problem at the start. Which signpost in that diagram would you use first?

Show the answer

The tangent at A meets two chords, AB and AD, at the point of contact. So the alternate segment theorem applies twice. The other signpost, four points on the circle, finishes the problem in Worked Example 5.

A method: label, give a reason, plan a route

Strong problem solvers do not wait for inspiration. They follow three habits, and the route usually appears while they do.

1. Label everything you know. Write every given angle on the diagram. Mark equal radii with small ticks. Put a right-angle mark wherever a radius meets a tangent, and wherever a point on the circle sees a diameter. Many routes stay invisible until the diagram is full.

2. Write a reason beside every new angle. Each time you find an angle, write the theorem that gave it. A reason is a test, not only a mark. If you cannot name the reason, you may be using a theorem whose trigger is not there.

3. Plan a route to the unknown. Start from the angle you want and work backwards. Ask: which theorem could give me this angle, and what would that theorem need? Then ask the same question about the angle it needs. Usually, two or three backward steps meet the facts you already have.

Mathematicians call this angle chasing. Each angle you find becomes a new fact that lets you find the next one, and the chain grows from both ends until they meet.

Stop and think

O is the centre of a circle, and A, B and C are on the circle. A student looks at the quadrilateral OACB and writes: $∠ A OB + ∠ A CB = 18 0^{\circ}$ , "opposite angles of a cyclic quadrilateral". Is this right?

Show the answer

No. A cyclic quadrilateral needs all four corners on the circle. The centre O is inside the circle, not on it. The correct link between the angle at O and the angle at C is the angle at the centre theorem: $∠ A OB = 2 \times ∠ A CB$ , when both stand on the same arc. The student's reason has no trigger, and writing it down is what exposes that.

Worked examples, from two steps to five

In every example, plan the route before you calculate, and end with a check by a different route.

Worked Example 1 — Two links

O is the centre of a circle. A, B and C are on the circle, and O lies inside triangle ABC. $∠ A BC = 3 2^{\circ}$ . Find $∠ A OC$ and $∠ O A C$ .

Plan the route. $∠ O A C$ is in triangle OAC, which is isosceles because OA and OC are radii. So you need its angle at O, and that comes from the 32° angle at B.

$∠ A OC = 2 \times 3 2^{\circ} = 6 4^{\circ}$ . Reason: the angle at the centre is twice the angle at the circumference, both standing on arc AC.

$∠ O A C = (18 0^{\circ} - 6 4^{\circ}) \div 2 = 5 8^{\circ}$ . Reason: OA = OC (radii), so triangle OAC is isosceles, and its angles add up to 180°.

Check. $6 4^{\circ} + 5 8^{\circ} + 5 8^{\circ} = 18 0^{\circ}$ . ✓ Notice also that $5 8^{\circ} = 9 0^{\circ} - 3 2^{\circ}$ . This is not luck: with the angle at B called $β$ , each base angle is $(18 0^{\circ} - 2 β) \div 2 = 9 0^{\circ} - β$ . You can rebuild this in one line, so there is nothing new to learn.

Worked Example 2 — A diameter and the same segment

AB is a diameter of a circle with centre O. C and D are on the same semicircle, in the order A, D, C, B. $∠ C A B = 2 5^{\circ}$ and $∠ D B A = 4 0^{\circ}$ . Find (a) $∠ A CB$ , (b) $∠ A BC$ , (c) $∠ CB D$ , (d) $∠ C A D$ , (e) $∠ A D C$ .

AB is a diameter. D and C are on the same semicircle. The angles at A and B are given.

Plan the route. Label first: the diameter gives right angles at C and D. After that, each part uses the part before it.

(a) $∠ A CB = 9 0^{\circ}$ . Reason: angle in a semicircle, because AB is a diameter.

(b) $∠ A BC = 18 0^{\circ} - 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ . Reason: angles in triangle ABC.

(d) $∠ C A D$ and $∠ CB D$ both stand on the chord CD, and A and B are on the same side of it. So $∠ C A D = ∠ CB D = 2 5^{\circ}$ . Reason: angles in the same segment are equal.

(e) ADCB is a cyclic quadrilateral, and D is opposite B. So $∠ A D C = 18 0^{\circ} - ∠ A BC = 18 0^{\circ} - 6 5^{\circ} = 11 5^{\circ}$ . Reason: opposite angles of a cyclic quadrilateral add up to 180°.

Check. Use a different route for triangle ACD. $∠ A C D$ and $∠ A B D$ both stand on the chord AD from the same side, so $∠ A C D = 4 0^{\circ}$ . Then the angles of triangle ACD are $2 5^{\circ} + 11 5^{\circ} + 4 0^{\circ} = 18 0^{\circ}$ .

Worked Example 3 — One tangent, two chords

XTY is the tangent to a circle at T. A and B are on the circle. $∠ A TX = 4 0^{\circ}$ and $∠ BT Y = 5 5^{\circ}$ . C is a point on the arc AB that does not contain T. Find $∠ A CB$ .

One tangent and two chords from T. The dashed line CT is added to split the unknown angle into two parts.

Plan the route. No theorem gives $∠ A CB$ directly from angles at T. But if you join C to T, the unknown splits into $∠ TC A$ and $∠ TCB$ . Each of these stands on a chord from T, so the alternate segment theorem can reach it. A line you add yourself, like CT, is called a construction line.

$∠ TC A = ∠ A TX = 4 0^{\circ}$ . Reason: alternate segment theorem for the chord TA. C is across the chord TA from the 40° angle.

$∠ TCB = ∠ BT Y = 5 5^{\circ}$ . Reason: alternate segment theorem for the chord TB. C is across the chord TB from the 55° angle.

$∠ A CB = 4 0^{\circ} + 5 5^{\circ} = 9 5^{\circ}$ .

Check. The angles at T on the straight line XTY add up to 180°, so $∠ A TB = 18 0^{\circ} - 4 0^{\circ} - 5 5^{\circ} = 8 5^{\circ}$ . TACB is a cyclic quadrilateral with T opposite C, so $∠ A CB = 18 0^{\circ} - 8 5^{\circ} = 9 5^{\circ}$ . ✓ Notice that the answer is obtuse. An obtuse angle at the circumference stands on the major arc, so C must be on the smaller arc AB.

Find the broken link

One wrong reason early in a chain spoils every line after it. The most common broken link in this topic is pairing an angle with the wrong angle at the centre. Find it in the working below.

Spot the error — a cyclic quadrilateral and its centre

ABCD is a cyclic quadrilateral with centre O. A is on the major arc BD, and C is on the minor arc BD. $∠ BO D = 14 0^{\circ}$ , the angle at O that is not reflex. A student finds $∠ B A D$ and $∠ BC D$ . Tap the line that contains the mistake.

The student's working has one wrong line. Here it is, with the correction.

∠BAD = ½ × 140° = 70°. Angle at the centre is twice the angle at the circumference, on arc BD. Correct: A is on the major arc, so ∠BAD stands on the minor arc BD, the same arc as the 140°.
∠BCD = ½ × 140° = 70°. This is the mistake. C is on the minor arc, so ∠BCD stands on the major arc BD. Its partner at the centre is the reflex angle BOD = 360° − 140° = 220°. So ∠BCD = 110°.
∠ABC + ∠ADC = 360° − 70° − 70° = 220°. The arithmetic is right, but it is built on the wrong line 2. With 110°, the sum is 180°.
Check: ∠BAD + ∠BCD = 70° + 70° = 140°. This should be 180°, because opposite angles of a cyclic quadrilateral add up to 180°. The failed check shows that a line above is wrong.

The student used the right theorem with the wrong partner. That is why a reason must name the arc, not only the theorem. And without the check, the wrong answer would have looked perfectly confident.

Worked Example 4 — Two tangents and the centre

Tangents from a point P touch a circle with centre O at A and B. $∠ A PB = 4 6^{\circ}$ . C is on the major arc AB and D is on the minor arc AB. Find $∠ A OB$ , $∠ O A B$ , $∠ A CB$ and $∠ A D B$ .

Two tangents from P. The radii OA and OB meet the tangents at right angles.

Plan the route. Label the two right angles at A and B first. Then quadrilateral OAPB gives the angle at O, and the rest follows from it.

$∠ O A P = ∠ OBP = 9 0^{\circ}$ . Reason: tangent is perpendicular to the radius.

$∠ A OB = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 4 6^{\circ} = 13 4^{\circ}$ . Reason: angles of quadrilateral OAPB add up to 360°.

$∠ O A B = (18 0^{\circ} - 13 4^{\circ}) \div 2 = 2 3^{\circ}$ . Reason: OA = OB (radii), so triangle OAB is isosceles.

$∠ A CB = 13 4^{\circ} \div 2 = 6 7^{\circ}$ . Reason: angle at the centre is twice the angle at the circumference, both on the minor arc AB.

$∠ A D B = 18 0^{\circ} - 6 7^{\circ} = 11 3^{\circ}$ . Reason: ACBD is a cyclic quadrilateral, and C is opposite D.

Check. Find $∠ A CB$ by a second route. PA = PB, so triangle PAB is isosceles and $∠ P A B = (18 0^{\circ} - 4 6^{\circ}) \div 2 = 6 7^{\circ}$ . This is the angle between the tangent PA and the chord AB, so the alternate segment theorem gives $∠ A CB = 6 7^{\circ}$ . ✓ The two routes agree.

Worked Example 5 — The problem from the start

Return to the first figure. ABCD is a cyclic quadrilateral, and XAY is the tangent at A. $∠ X A B = 6 5^{\circ}$ and $∠ D A Y = 4 8^{\circ}$ . Find $∠ BC D$ .

Plan the route. Work backwards from the unknown. $∠ BC D$ is a corner of the cyclic quadrilateral, and the opposite corner is A. So $∠ BC D = 18 0^{\circ} - ∠ B A D$ . Now the question has become: what is $∠ B A D$ ? It is the third angle on the straight line at A.

$∠ A D B = ∠ X A B = 6 5^{\circ}$ . Reason: alternate segment theorem for the chord AB. D is across the chord AB from X.

$∠ A B D = ∠ D A Y = 4 8^{\circ}$ . Reason: alternate segment theorem for the chord AD. B is across the chord AD from Y.

$∠ B A D = 18 0^{\circ} - 6 5^{\circ} - 4 8^{\circ} = 6 7^{\circ}$ . Reason: angles on the straight line XAY.

$∠ BC D = 18 0^{\circ} - 6 7^{\circ} = 11 3^{\circ}$ . Reason: opposite angles of a cyclic quadrilateral add up to 180°.

Check. In triangle ABD, $6 5^{\circ} + 4 8^{\circ} + 6 7^{\circ} = 18 0^{\circ}$ . ✓ The angle at A was found from the straight line, and the other two from the tangent, so this check really is a second route.

Now look at the answer again: $11 3^{\circ} = 6 5^{\circ} + 4 8^{\circ}$ . The straight line gives $∠ B A D = 18 0^{\circ} - (6 5^{\circ} + 4 8^{\circ})$ , and the cyclic quadrilateral gives $∠ BC D = 18 0^{\circ} - ∠ B A D$ . The two 180°s cancel. So in every figure of this shape, the angle at C is the two tangent-chord angles added together.

That is it. That is the whole skill. Each step used one theorem with one reason, and the route came from asking, again and again, "what would give me this angle?"

Three traps

Trap 1 — Trusting how the diagram looks

Exam diagrams are often not drawn to scale. A chord that looks like a diameter, or a line that looks like a tangent, cannot be used unless the question says so or you can prove it.

Trap 2 — A theorem without its trigger

The most common case is a quadrilateral with a corner at the centre, treated as cyclic. Another is the alternate segment theorem used for a chord that does not start at the point of contact. Before you write a reason, check that its trigger is really in the diagram.

Trap 3 — A reason that names no arc

"Circle theorem" is not a reason, and examiners give it no marks. Even the correct name can hide the wrong partner, as the spot-the-error working showed. Name the arc or chord the angle stands on, and the mistake becomes easy to see.

Try it yourself

Work each one on paper first. Draw a large diagram, label everything you know, and write a reason beside every angle. Open a hint only when you are stuck.

A, B, C and D lie on a circle with centre O, in that order. $∠ B A D = 7 8^{\circ}$ and $∠ A D B = 4 0^{\circ}$ . Find (a) $∠ BC D$ , (b) $∠ A CB$ , (c) $∠ BO D$ , the angle at O that stands on the arc BCD.

Hint 1
Look for a cyclic quadrilateral, a pair of angles on the same chord, and an angle at the centre.

Hint 2
$∠ A CB$ and $∠ A D B$ both stand on the chord AB. Are C and D on the same side of it?

Full answer
(a) $∠ BC D = 18 0^{\circ} - 7 8^{\circ} = 10 2^{\circ}$ (opposite angles of a cyclic quadrilateral). (b) C and D are on the same side of AB, so $∠ A CB = ∠ A D B = 4 0^{\circ}$ (angles in the same segment). (c) $∠ B A D$ stands on the arc BCD, so $∠ BO D = 2 \times 7 8^{\circ} = 15 6^{\circ}$ (angle at the centre is twice the angle at the circumference). Check: the reflex angle BOD is $36 0^{\circ} - 15 6^{\circ} = 20 4^{\circ}$ , and half of it is $10 2^{\circ} = ∠ BC D$ , which stands on the other arc. ✓
AB is a diameter of a circle, and C is on the circle with $∠ B A C = 2 8^{\circ}$ . The tangent at C meets the line AB, extended beyond B, at P. Find $∠ BCP$ , $∠ A BC$ and $∠ CPB$ .

Hint 1
The tangent at C and the chord CB meet at the point of contact. Which theorem does that switch on?

Hint 2
$∠ A BC$ is an exterior angle of triangle BCP. It equals the two interior opposite angles added together.

Full answer
$∠ BCP = ∠ B A C = 2 8^{\circ}$ (alternate segment theorem for the chord CB; A is across the chord from P). $∠ A CB = 9 0^{\circ}$ (angle in a semicircle), so $∠ A BC = 18 0^{\circ} - 9 0^{\circ} - 2 8^{\circ} = 6 2^{\circ}$ (angles in triangle ABC). $∠ A BC$ is the exterior angle of triangle BCP at B, so $∠ CPB = 6 2^{\circ} - 2 8^{\circ} = 3 4^{\circ}$ . Check with the centre O: $∠ COB = 2 \times 2 8^{\circ} = 5 6^{\circ}$ and $∠ OCP = 9 0^{\circ}$ (tangent and radius), so in triangle OCP the angle at P is $18 0^{\circ} - 9 0^{\circ} - 5 6^{\circ} = 3 4^{\circ}$ . ✓
Tangents from a point P touch a circle at A and B. C is a point on the major arc AB, and $∠ A CB = 5 8^{\circ}$ . Find $∠ A PB$ .

Hint 1
This time the angle at the circumference is given, and the angle at P is wanted. Work backwards: which angle at A would give you the triangle PAB?

Hint 2
$∠ P A B$ is the angle between the tangent PA and the chord AB. The alternate segment theorem links it to $∠ A CB$ .

Full answer
$∠ P A B = ∠ A CB = 5 8^{\circ}$ (alternate segment theorem). PA = PB (tangents from a point), so $∠ PB A = 5 8^{\circ}$ too. So $∠ A PB = 18 0^{\circ} - 5 8^{\circ} - 5 8^{\circ} = 6 4^{\circ}$ (angles in triangle PAB). Check: $∠ A OB = 2 \times 5 8^{\circ} = 11 6^{\circ}$ , and in quadrilateral OAPB, $11 6^{\circ} + 9 0^{\circ} + 9 0^{\circ} + 6 4^{\circ} = 36 0^{\circ}$ . ✓
The tangent at T is XTY. A and B are on the circle, with A on the X side. A line parallel to XY cuts TA at P and TB at Q, between T and the circle. Prove that A, P, Q and B lie on one circle. Draw the chords TA and TB first, then a line across them, parallel to the tangent.

Hint 1
To prove that four points lie on a circle, use the converse of the cyclic quadrilateral theorem: show that a pair of opposite angles of APQB adds up to 180°.

Hint 2
Call $∠ A TX = α$ . Find $∠ TPQ$ from the parallel lines, and $∠ TB A$ from the alternate segment theorem. Both are $α$ .

Full answer
Let $∠ A TX = α$ . PQ is parallel to XY, so $∠ TPQ = ∠ A TX = α$ (alternate angles, parallel lines). So $∠ A PQ = 18 0^{\circ} - α$ (angles on the straight line TPA). By the alternate segment theorem for the chord TA, $∠ TB A = α$ , and this is the angle of APQB at B, $∠ A BQ$ . So $∠ A PQ + ∠ A BQ = (18 0^{\circ} - α) + α = 18 0^{\circ}$ . A pair of opposite angles of APQB adds up to 180°, so APQB is a cyclic quadrilateral (converse of the cyclic quadrilateral theorem). Check with numbers: if $α = 3 0^{\circ}$ , then $∠ A PQ = 15 0^{\circ}$ and $∠ A BQ = 3 0^{\circ}$ , which add up to 180°. ✓ This is the stretch: two circle theorems, parallel lines and a converse, in one chain.

The bigger idea

Look back at the toolkit table. It has seven rows, but it does not hold seven separate ideas. Angles in the same segment, the angle in a semicircle and the cyclic quadrilateral are all children of one theorem: the angle at the centre is twice the angle at the circumference. Equal tangents and the alternate segment theorem both rest on one more fact: the tangent is perpendicular to the radius. So a five-step exam problem really uses two ideas, again and again.

This is how mathematics is built. Euclid's Elements, written in Greek about 2300 years ago, proves each result from results proved before it. Most of the circle theorems you used are among Propositions 18 to 32 of its third book. Every problem in this lecture is a small copy of that structure: a chain of reasons from what is given to what is wanted. Later lectures prove geometric results with vectors, by a completely different route, and the same skill carries over.

Key concepts in this lesson

An exam diagram hides several theorems, and the skill is to find them and connect them into a chain, one theorem per step.
Each theorem has a trigger you can see: a centre, a diameter, four points on the circle, a radius to a tangent, two tangents from a point, or a tangent meeting a chord at its point of contact.
Label everything first, write a reason beside every new angle, and plan the route by working backwards from the unknown.
A reason should name the arc or chord that an angle stands on, because the most common error is pairing an angle with the wrong partner.
A check by a second route, such as the angles of a triangle or a second theorem, shows whether every link in the chain is sound.