Angle at the centre — the key circle theorem — Mathematics

Before this

Parts of a circle · coming soon
Properties of triangles · coming soon
Angles on lines and at points · coming soon

This leads to

Three friends and one gate

A garden is a perfect circle, with a fence all the way round. In one part of the fence there is a gate between two posts, A and B. Three friends, Mateo, Kenji and Lina, stand at different places on the fence, across the garden from the gate. Each one measures the angle between the line to post A and the line to post B.

Mateo stands much closer to the gate than the others. Kenji stands exactly opposite it. Who sees the gate at the widest angle?

Most people answer "Mateo". A door close to you fills more of your view than a door far away. So the angle should grow as you get closer.

On this fence, it does not. All three friends measure exactly the same angle: 40°. Move any of them anywhere along that part of the fence, and the angle is still 40°.

The gate AB, seen from three places on the circle (M for Mateo, K for Kenji, L for Lina). Every one of them sees it at 40°. From the centre O, it is seen at 80°.

There is a second surprise. From the centre of the garden, O, the gate is seen at 80°: exactly twice what every friend measured.

This is not a coincidence. It is true for every circle and every gate, and by the end of this lecture you will see why it must be true. The reason uses only two facts you already know: all radii of a circle are equal, and the angles of a triangle add up to 180°.

Naming the parts

Put two points, A and B, on a circle with centre O. They cut the circle into two arcs. The shorter one is the minor arc. The longer one is the major arc. In the garden, the gate is the minor arc, and the friends stand on the major arc.

Now draw two kinds of angle. The angle at the centre, $∠ A OB$ , has its corner at O and the radii OA and OB as its arms. The angle at the circumference, $∠ A PB$ , has its corner at another point P on the circle's edge (the circumference).

Both angles have arms that end at A and at B. We say that both angles stand on the arc AB: the arc that lies across from the corner, between the ends of the two arms. Some books say that the arc AB "subtends" the angle. It means the same thing.

In a three-letter name like $∠ A PB$ , the middle letter is always the corner.

Watch the angle refuse to change

Test the claim yourself before any proof. In the explorer below, the thick arc is the arc that $∠ A PB$ stands on.

Explore — move P around the circle

Move P with the slider or the − / + buttons. First keep P on the major arc (the long way round). Then choose a different size for the angle at the centre. Last, move P onto the minor arc, between A and B, and read what happens.

Angle at the centre:

Position of P around the circle: 90°

This table shows the angle at P for four positions of P, on a circle where the angle at the centre on the minor arc AB is 100°.

Where P is	∠APB	Angle at the centre on the same arc
Major arc, top of the circle	50°	100°
Major arc, near A	50°	100°
Major arc, near B	50°	100°
Minor arc, between A and B	130°	260° (reflex)

Here is what you should have seen. While P stays on the major arc, $∠ A PB$ does not change at all, wherever P goes. It is always exactly half of $∠ A OB$ . When you change the size of the angle at the centre, the angle at P changes too, but it is still exactly half.

On the minor arc, the angle at P suddenly becomes obtuse. Now the arc across from P is the major arc, and the angle at the centre on that arc is the reflex angle AOB, the one bigger than 180°. The angle at P is exactly half of it. The rule did not break. It changed partners.

Stop and think

Mateo walks along the fence towards post A. The line from him to A gets shorter and shorter. Just before he reaches A, is his angle almost zero?

Show the answer

No. It is still 40°, right up to the moment he reaches A. An angle measures the difference in direction between two lines, not their length. As Mateo gets close to A, the line to A becomes very short, but its direction settles down: it points along the edge of the circle. Only when he stands exactly on A is there no angle, because there is only one line. The alternate segment theorem lecture uses exactly this idea, because that edge direction is the tangent.

Why it must be true

The explorer tests some positions, not every position. For every position we need a proof: one clever move, then two facts.

The clever move. Draw a straight line from P, through the centre O, to the other side of the circle. Call the point where it meets the circle Q. Now look at the picture. The line PO cuts the diagram into two triangles, OAP and OBP. Each of these triangles has two sides that are radii.

The line from P through O to Q. Triangles OAP and OBP are isosceles, with base angles α and β. The angles at O are 2α and 2β.

The first fact: equal radii make isosceles triangles. OA and OP are both radii, so they are equal. This means triangle OAP is isosceles. The two angles opposite the equal sides are equal, so $∠ O A P = ∠ OP A$ . Call each of them α (the Greek letter alpha). In the same way, OB = OP, so triangle OBP is isosceles, with two angles β (beta).

The second fact: an exterior angle of a triangle equals the two interior opposite angles added together. This comes straight from 180°. In triangle OAP, the three angles add up to 180°, so $∠ A OP = 18 0^{\circ} - 2 α$ . The line POQ is straight, so the angles on it at O also add up to 180°. This means $∠ A OQ = 18 0^{\circ} - ∠ A OP = 2 α$ . The angle at the centre on this side is double the base angle.

Now put the pieces together, one step at a time.

Step through — proving the angle at the centre theorem

Use the figure above. Press Next to add one line of the proof. Read the reason before you go on.

Each line of the proof, with the reason it is allowed.

$O A = OP = OB$ . All three are radii of the same circle.
$∠ O A P = ∠ OP A = α$ . Triangle OAP is isosceles, so its base angles are equal.
$∠ A OQ = 2 α$ . It is the exterior angle of triangle OAP at O.
$∠ OBP = ∠ OPB = β$ . Triangle OBP is isosceles too.
$∠ BOQ = 2 β$ . The same reason, in the second triangle.
$∠ A OB = 2 α + 2 β = 2 (α + β)$ . The line OQ splits $∠ A OB$ into two parts.
$∠ A PB = α + β$ . The line PO splits $∠ A PB$ into two parts.

So $∠ A OB = 2 \times ∠ A PB$ , for every P on the major arc with O inside angle APB.

That is it. That is the whole idea. Equal radii make isosceles triangles, and each isosceles triangle doubles its base angle at the centre. The two halves of the angle at P become the two halves of the angle at O, each one doubled.

Notice that the proof never needed to know where exactly P is. That is why the angle at P cannot change as P moves: it is always half of the same $∠ A OB$ .

When the centre is outside the angle

There is one situation the proof above does not cover. Sometimes A and B are both on the same side of the line PQ. Then O is outside the angle APB, and the line PQ does not split the angles into two parts.

Here A and B are on the same side of the line PQ, so O is outside angle APB. The angle at O (40°) is still twice the angle at P (20°).

The same two facts still work; we subtract instead of add. Let $∠ OP A = α$ and $∠ OPB = β$ . Triangles OAP and OBP are isosceles, exactly as before. So $∠ A OQ = 2 α$ and $∠ BOQ = 2 β$ . In this picture, B is closer to Q than A is, so both angles at P and both angles at O are differences:

∠ A PB = α - β ∠ A OB = 2 α - 2 β = 2 (α - β)

So once again $∠ A OB = 2 \times ∠ A PB$ . In the figure, $α = 5 5^{\circ}$ and $β = 3 5^{\circ}$ . This gives $∠ A PB = 2 0^{\circ}$ and $∠ A OB = 4 0^{\circ}$ .

This proof is very old. It is Proposition 20 in Book III of Euclid's Elements, written about 2300 years ago. Euclid used the same line through the centre, and treated the second case by subtraction.

The angle at the centre theorem

The angle at the centre of a circle is twice the angle at the circumference, when both angles stand on the same arc.

$∠ A OB = 2 \times ∠ A PB$

If P is on the minor arc, the angle at the centre on the same arc is the reflex angle AOB.

When the angle at the circumference is obtuse

Go back to the explorer's last surprise. Suppose A and B are close together near the top of a circle, and P is on the small arc between them. Then $∠ A PB$ opens downwards, towards the large part of the circle. The arc it stands on is the major arc.

P is on the minor arc, so ∠APB = 125° stands on the major arc (thick line). The angle at the centre on that arc is reflex ∠AOB = 250°. The grey point R on the major arc sees AB at 55°.

The angle at the centre that stands on the major arc is the reflex angle at O. The same proof works for it. The line from P through O now reaches Q on the major arc, and the two angles 2α and 2β at O together make the reflex angle. So reflex $∠ A OB = 2 (α + β) = 2 \times ∠ A PB = 25 0^{\circ}$ .

A quick test tells you which partner to use. An acute angle at the circumference pairs with an angle at the centre less than 180°. An obtuse one pairs with a reflex angle, because double an angle more than 90° is more than 180°.

Stop and think

In the figure, $∠ A PB = 12 5^{\circ}$ , and the ordinary (non-reflex) angle $∠ A OB$ is $36 0^{\circ} - 25 0^{\circ} = 11 0^{\circ}$ . A student writes " $∠ A PB = 11 0^{\circ} \div 2 = 5 5^{\circ}$ ". What went wrong?

Show the answer

The 110° angle stands on the minor arc. But P is on the minor arc, so $∠ A PB$ stands on the major arc. The student paired two angles on different arcs. The 55° is real, but it belongs to any point R on the major arc, like the grey point in the figure. Notice that $12 5^{\circ} + 5 5^{\circ} = 18 0^{\circ}$ . The next lecture shows why that always happens.

A free result: angles on the same arc are equal

Now the garden puzzle has its answer. Mateo, Kenji and Lina all stand on the major arc, so each of their angles stands on the gate, the minor arc AB. Each angle is half of $∠ A OB = 8 0^{\circ}$ , so each is 40°. They must be equal, because they are all half of the same thing.

This is the second circle theorem, and it came free from the first. Angles at the circumference that stand on the same arc are equal. Many books call this "angles in the same segment are equal". A segment is the region between a chord and its arc, and the friends all stand in the same segment. Euclid proves it in the very next proposition, Book III Proposition 21, using Proposition 20.

There is one condition: the points must be on the same side of the chord AB. That is why R in the last figure sees 55°, while P sees 125°.

Worked examples

In every example, O is the centre of the circle, and A, B and P are points on the circle.

Worked Example 1 — From the circumference to the centre

P is on the major arc and $∠ A PB = 3 5^{\circ}$ . Find $∠ A OB$ .

First ask: do both angles stand on the same arc? $∠ A PB$ has arms ending at A and B, and P is on the major arc. So it stands on the minor arc. So does $∠ A OB$ . The theorem applies.

∠ A OB = 2 \times 3 5^{\circ} = 7 0^{\circ}

Write the reason in words: "The angle at the centre is twice the angle at the circumference." In an exam, the reason earns marks.

Check: $∠ A PB$ is acute, so the angle at the centre should be less than 180°. 70° is.

Worked Example 2 — From the centre to the circumference, and one step more

$∠ A OB = 11 2^{\circ}$ , and P is on the major arc. Find $∠ A PB$ and $∠ O A B$ .

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:

∠ A PB = 11 2^{\circ} \div 2 = 5 6^{\circ}

For $∠ O A B$ , look at triangle OAB. OA and OB are radii, so the triangle is isosceles and its base angles are equal:

∠ O A B = \frac{18 0 ^{\circ} - 11 2 ^{\circ}}{2} = 3 4^{\circ}

Check: $3 4^{\circ} + 3 4^{\circ} + 11 2^{\circ} = 18 0^{\circ}$ .

Worked Example 3 — Using the proof's own idea

O is inside angle APB. You are told that $∠ O A P = 2 5^{\circ}$ and $∠ OBP = 3 0^{\circ}$ . Find $∠ A OB$ .

Join OP. Triangle OAP is isosceles (OA = OP), so $∠ OP A = 2 5^{\circ}$ . Triangle OBP is isosceles (OB = OP), so $∠ OPB = 3 0^{\circ}$ . The line OP splits the angle at P, so

∠ A PB = 2 5^{\circ} + 3 0^{\circ} = 5 5^{\circ}, ∠ A OB = 2 \times 5 5^{\circ} = 11 0^{\circ}

Check: the shape OAPB has four corners, so its angles add up to 360°. Its angle at O is the reflex angle, $36 0^{\circ} - 11 0^{\circ} = 25 0^{\circ}$ , because O is a "dent" in the shape. Then $2 5^{\circ} + 5 5^{\circ} + 3 0^{\circ} + 25 0^{\circ} = 36 0^{\circ}$ . ✓

Three traps

Trap 1 — An angle that stands on a different arc

The theorem connects two angles only when the arms of both end at the same two points. In a busy diagram, $∠ A OB$ and an angle at P whose arms go to A and C stand on different arcs, and the theorem says nothing about them. Before you double or halve, trace both pairs of arms with your finger.

Trap 2 — Pairing an obtuse angle with the wrong angle at the centre

If the angle at the circumference is obtuse, its partner at the centre is reflex. A student who halves the ordinary angle at O gets the angle at a point on the other arc. The test from earlier catches this: acute pairs with less than 180°, obtuse pairs with more than 180°.

Trap 3 — A point that only looks like the centre

The proof depends on OA, OB and OP being equal, which is only true when O is the centre. A point where two chords cross may look central, but the theorem does not apply there. Use it only when the question says the point is the centre. And write the reason in full ("the angle at the centre is twice the angle at the circumference"), not just "circle theorem".

Try it yourself

Work each one on paper first. Draw a diagram for every question. Open a hint only when you are stuck.

O is the centre of a circle, and P is on the major arc AB. $∠ A OB = 14 8^{\circ}$ . Find $∠ A PB$ and $∠ O A B$ .

Hint 1
Both angles stand on the minor arc, so one is half the other. For $∠ O A B$ , look at triangle OAB.

Hint 2
OA = OB, so triangle OAB is isosceles. Its two base angles share what is left of 180° after the 148°.

Full answer
$∠ A PB = 14 8^{\circ} \div 2 = 7 4^{\circ}$ (the angle at the centre is twice the angle at the circumference). In triangle OAB, OA = OB, so $∠ O A B = (18 0^{\circ} - 14 8^{\circ}) \div 2 = 1 6^{\circ}$ . Check: $1 6^{\circ} + 1 6^{\circ} + 14 8^{\circ} = 18 0^{\circ}$ , and 74° is acute, which matches an angle at the centre less than 180°.
P is on the major arc AB. $∠ A OB = (5 x - 10)^{\circ}$ and $∠ A PB = (2 x + 7)^{\circ}$ . Find x, and both angles.

Hint 1
The theorem gives you an equation: the angle at the centre equals 2 times the angle at the circumference.

Hint 2
Write $5 x - 10 = 2 (2 x + 7)$ , then expand the bracket.

Full answer
$5 x - 10 = 2 (2 x + 7) = 4 x + 14$ . Subtract 4x from both sides: $x - 10 = 14$ , so $x = 24$ . Then $∠ A OB = 5 \times 24 - 10 = 11 0^{\circ}$ and $∠ A PB = 2 \times 24 + 7 = 5 5^{\circ}$ . Check: $2 \times 5 5^{\circ} = 11 0^{\circ}$ . ✓
P is on the minor arc AB, and $∠ A PB = 10 8^{\circ}$ . Find the reflex angle AOB, the ordinary angle AOB, and $∠ O A B$ .

Hint 1
108° is obtuse. Which arc does $∠ A PB$ stand on, and which angle at O stands on the same arc?

Hint 2
Double 108° to get the reflex angle. The ordinary angle AOB is what is left of 360°. Then use the isosceles triangle OAB.

Full answer
P is on the minor arc, so $∠ A PB$ stands on the major arc, and its partner is the reflex angle: reflex $∠ A OB = 2 \times 10 8^{\circ} = 21 6^{\circ}$ . The ordinary $∠ A OB = 36 0^{\circ} - 21 6^{\circ} = 14 4^{\circ}$ . In the isosceles triangle OAB, $∠ O A B = (18 0^{\circ} - 14 4^{\circ}) \div 2 = 1 8^{\circ}$ . Check: a point on the major arc would see AB at $14 4^{\circ} \div 2 = 7 2^{\circ}$ , and $10 8^{\circ} + 7 2^{\circ} = 18 0^{\circ}$ , the pattern from the Stop-and-think.
A regular pentagon ABCDE is drawn inside a circle with centre O, so that all five corners are on the circle. Use the angle at the centre theorem to find (a) $∠ A OB$ , (b) $∠ A CB$ , and (c) the interior angle $∠ A BC$ of the pentagon. Then check (c) another way.

Hint 1
The five sides are equal, so the five angles at the centre that stand on them are equal. They go all the way round O.

Hint 2
For (c), ask which arc $∠ A BC$ stands on. Its arms end at A and C, and the arc across from B goes A, E, D, C. That arc covers three sides of the pentagon.

Full answer
(a) The five equal angles at O add up to 360°, so $∠ A OB = 36 0^{\circ} \div 5 = 7 2^{\circ}$ .
(b) C is on the major arc AB, so $∠ A CB$ stands on the minor arc AB: $∠ A CB = 7 2^{\circ} \div 2 = 3 6^{\circ}$ . ( $∠ A D B$ and $∠ A EB$ are also 36°, because they stand on the same arc.)
(c) $∠ A BC$ stands on the arc from A through E and D to C. That arc covers three sides, so the angle at the centre on it is the reflex angle $3 \times 7 2^{\circ} = 21 6^{\circ}$ . So $∠ A BC = 21 6^{\circ} \div 2 = 10 8^{\circ}$ .
Check: the interior angles of any pentagon add up to $(5 - 2) \times 18 0^{\circ} = 54 0^{\circ}$ , so each angle of a regular pentagon is $54 0^{\circ} \div 5 = 10 8^{\circ}$ . ✓ Two completely different methods give the same answer, and the second one needed the reflex case.

The bigger idea

This one theorem is the parent of most of the circle theorems you will meet. The equal angles in the same segment came from it in one line. The next lecture takes the special case where A and B are at the two ends of a diameter, so that the angle at the centre is a straight line, 180°. It also explains the pattern you saw twice in this lecture, two angles adding up to 180°.

There is a deeper reason this theorem matters. The angle at the centre measures the arc itself: double the angle at the centre and the arc doubles in length. This is why some books, especially in the United States, say "an inscribed angle is half of the arc it cuts off", with the arc measured in degrees. Later, when you measure angles in radians, the angle at the centre becomes the length of the arc on a circle of radius 1.

Key concepts in this lesson

The angle at the centre of a circle is twice the angle at the circumference when both stand on the same arc.
An angle stands on the arc that lies across from its corner, between the ends of its two arms.
The proof draws a line from P through the centre and uses two facts: equal radii make isosceles triangles, and an exterior angle equals the two interior opposite angles added together.
An obtuse angle at the circumference stands on the major arc, and its partner at the centre is the reflex angle.
Angles at the circumference that stand on the same arc are equal, because each one is half of the same angle at the centre.