Theorems on Central and Inscribed Angles
The same chord or arc of a circle can form angles at the centre and at different points on the circumference. Some fixed relationships always exist between these angles. Once you understand these relationships, many circle proofs, unknown-angle problems and questions on concyclic points become much easier.
This page explains four important theorems about angles in a circle and shows how to use them.
📚 First, understand a few basic ideas
👉 Central Angle: An angle whose vertex is at the centre of the circle. For example, ∠AOB.
👉 Inscribed Angle: An angle whose vertex lies on the circumference and whose two arms meet the circle. For example, ∠ACB.
👉 Angles standing on the same arc: If the arms of ∠AOB and ∠ACB meet the circle at the same points A and B, both angles stand on the same arc AB.
👉 Same segment: Points lying on the same side of a chord and on the same part of the circumference are said to lie in the same segment for this theorem.
⭐ Theorem 1: The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining part of the circle
| Given: | In a circle with centre O, the arc AB subtends the central angle ∠AOB and the inscribed angle ∠ACB. |
| To prove: | ∠AOB = 2∠ACB. |
| Construction: | Join C to O and extend CO beyond O to a point D. |
| Proof: | In △AOC, OA = OC. [Radii of the same circle] |
| ∴ | ∠OAC = ∠ACO. [Base angles of an isosceles triangle are equal] |
| Also, | Since CO is extended to D, the exterior angle of △AOC gives ∠AOD = ∠OAC + ∠ACO. |
| ∴ | ∠AOD = 2∠ACO. ...(i) |
| Similarly, | In △BOC, OB = OC. [Radii of the same circle] |
| ∴ | ∠OBC = ∠BCO. |
| And, | The exterior angle ∠BOD = ∠OBC + ∠BCO = 2∠BCO. ...(ii) |
| Now, | If the centre O lies inside ∠ACB, then ∠AOB = ∠AOD + ∠DOB and ∠ACB = ∠ACO + ∠OCB. |
| ∴ | From (i) and (ii), ∠AOB = 2∠ACO + 2∠OCB = 2(∠ACO + ∠OCB). |
| ∴ | ∠AOB = 2∠ACB. [Proved] |
📚 The relation is true in the other positions too
The centre O may lie on one arm of ∠ACB or outside the angle. In those cases, the required angle relations involve subtraction instead of addition, but the same result is obtained:
∠AOB = 2∠ACB.
So, whether the relevant arc is a minor arc, a semicircle or a major arc, the appropriate central angle is always twice the inscribed angle standing on the same arc.
📚 Direct meaning of the theorem
For the same arc,
Central angle = 2 × Inscribed angle.
Conversely,
Inscribed angle = 1/2 × Central angle.
💡 Simple example
If the central angle standing on an arc is 100°, the inscribed angle standing on the same arc is
100° ÷ 2 = 50°.
⭐ Theorem 2: Angles in the same segment of a circle are equal
| Given: | In a circle with centre O, points C and D lie in the same segment of chord AB. The angles ∠ACB and ∠ADB stand on the same chord or arc AB. |
| To prove: | ∠ACB = ∠ADB. |
| Construction: | Join OA and OB. |
| Proof: | The central angle ∠AOB and the inscribed angle ∠ACB stand on the same arc AB. |
| ∴ | By Theorem 1, ∠AOB = 2∠ACB. ...(i) |
| Again, | The same central angle ∠AOB and the inscribed angle ∠ADB stand on arc AB. |
| ∴ | By Theorem 1, ∠AOB = 2∠ADB. ...(ii) |
| ∴ | From (i) and (ii), 2∠ACB = 2∠ADB. |
| Therefore, | ∠ACB = ∠ADB. [Proved] |
📚 Simple meaning of the theorem
No matter which point in the same segment of a circle is used to view the same chord AB, the angle formed by that chord is the same.
💡 Simple example
If C and D lie in the same segment of chord AB and ∠ACB = 42°, then
∠ADB = 42°.
⭐ Theorem 3: If a line segment subtends equal angles at two points on the same side of it, then the four points are concyclic
This is an important converse of the theorem on angles in the same segment.
| Given: | Points C and D lie on the same side of segment AB and ∠ACB = ∠ADB. |
| To prove: | The four points A, B, C, D are concyclic. |
| Construction: | Draw the circle through the three non-collinear points A, B, C. Let line AD meet this circle again at D′. Join BD′. |
| Proof: | Points A, B, C, D′ lie on the same circle. Therefore, by the theorem on angles in the same segment, |
| ∴ | ∠AD′B = ∠ACB. ...(i) |
| Again, | It is given that ∠ADB = ∠ACB. ...(ii) |
| ∴ | From (i) and (ii), ∠AD′B = ∠ADB. |
| But, | A, D, D′ are collinear and D and D′ lie on the same side of segment AB. From a point on the same side of a fixed line, a ray making a fixed angle with that line has a unique direction. |
| ∴ | DB and D′B lie on the same straight line. Also, both D and D′ lie on line AD. |
| Therefore, | D ≡ D′. Hence D also lies on the circle through A, B, C. |
| Thus, | A, B, C, D are concyclic. [Proved] |
📚 A quick test for concyclic points
If C and D lie on the same side of segment AB and
∠ACB = ∠ADB,
then we can conclude that
A, B, C, D are concyclic.
⭐ Theorem 4: The angle in a semicircle is a right angle
| Given: | In a circle with centre O, AB is a diameter and C is any point on the semicircle. Thus ∠ACB is an angle in a semicircle. |
| To prove: | ∠ACB = 90°. |
| Proof: | Since AB is a diameter, the straight angle at the centre is ∠AOB = 180°. |
| Again, | The central angle ∠AOB and the inscribed angle ∠ACB stand on the same semicircular arc AB. |
| ∴ | By Theorem 1, ∠AOB = 2∠ACB. |
| ∴ | 180° = 2∠ACB. |
| ∴ | ∠ACB = 90°. |
| Therefore, | The angle in a semicircle is a right angle. [Proved] |
📚 Another proof
Join OC. Since OA = OC and OB = OC, both △AOC and △BOC are isosceles triangles.
Let
∠OAC = ∠OCA = x and ∠OBC = ∠OCB = y.
Then,
∠ACB = x + y.
In △ABC,
∠BAC + ∠ABC + ∠ACB = 180°.
Therefore,
x + y + (x + y) = 180°.
So,
2(x + y) = 180°,
and hence,
∠ACB = x + y = 90°.
📚 Application 1: Finding a central angle from an inscribed angle
If an inscribed angle standing on an arc is 35°, then
Central angle = 2 × 35° = 70°.
Therefore, the central angle is 70°.
📚 Application 2: Finding an inscribed angle from a central angle
If the central angle standing on an arc is 124°, then
Inscribed angle = 124° ÷ 2 = 62°.
📚 Application 3: Finding an unknown angle in the same segment
Suppose C and D lie in the same segment of chord AB and ∠ACB = 48°.
By the theorem on angles in the same segment,
∠ADB = ∠ACB = 48°.
📚 Application 4: Checking whether four points are concyclic
Suppose C and D lie on the same side of segment AB, and
∠ACB = 58°, ∠ADB = 58°.
Hence,
∠ACB = ∠ADB.
Therefore, by the converse theorem, A, B, C, D are concyclic.
📚 Application 5: Recognising a right-angled triangle using a diameter
Suppose AB is a diameter of a circle and C is a point on the circle.
By the angle-in-a-semicircle theorem,
∠ACB = 90°.
Therefore, △ABC is a right-angled triangle and AB is its hypotenuse.
This result is very useful for identifying right-angled triangles inside a circle.
🧩 NCERT-style solved examples
Example 1
In a circle with centre O, ∠AOB = 146°. Point C lies on the remaining part of the circle and ∠ACB stands on the same arc AB. Find ∠ACB.
Solution:
∠ACB = 1/2 × ∠AOB = 1/2 × 146° = 73°.
So, ∠ACB = 73°.
Example 2
Points P and Q lie in the same segment of chord AB. If ∠APB = 3x + 5° and ∠AQB = 5x - 25°, find x and the common angle.
Solution:
Angles in the same segment are equal.
3x + 5 = 5x - 25 2x = 30x = 15.
Therefore,
∠APB = 3 × 15 + 5 = 50°.
Hence both angles are 50°.
Example 3
AB is a diameter of a circle. Point C lies on the circle. If ∠CAB = 32°, find ∠ABC.
Solution:
Since AB is a diameter,
∠ACB = 90°.
Using the angle sum of △ABC,
∠ABC = 180° - 90° - 32° = 58°.
✍️ Practice problems
-
The angle subtended by an arc at the centre is 132°. Find the angle subtended by the same arc at a point on the remaining part of the circle.
-
An inscribed angle standing on arc PQ is 47°. Find the central angle standing on the same arc.
-
Points C and D lie in the same segment of chord AB. If ∠ACB = 64°, find ∠ADB.
-
Points C and D lie on the same side of AB. If ∠ACB = 71° and ∠ADB = 71°, what can you say about A, B, C, D?
-
AB is a diameter of a circle and C is a point on the circle. If ∠BAC = 41°, find ∠ABC.
-
In the same segment of a circle, ∠APB = 4x + 8° and ∠AQB = 6x - 22°. Find x and the two angles.
Answers
- 66°
- 94°
- 64°
- A, B, C, D are concyclic
- 49°
- x = 15, and each angle is 68°
📚 The four theorems at a glance
| Topic | Main result |
|---|---|
| Central and inscribed angles | Central angle = 2 × Inscribed angle |
| Angles in the same segment | All inscribed angles in the same segment standing on the same chord are equal |
| Condition for concyclic points | ∠ACB = ∠ADB ⇒ A, B, C, D are concyclic [same side] |
| Angle in a semicircle | 90° |
⚠️ Common mistakes
👉 Wrong: Any central angle is twice any inscribed angle.
Correct: The two angles must stand on the same arc.
👉 Wrong: All inscribed angles standing on the same chord are equal.
Correct: Their vertices must lie in the same segment, that is, on the same side of the chord on the same part of the circle.
👉 Wrong: If ∠ACB = ∠ADB, the four points are always concyclic.
Correct: When using this form of the converse theorem, check that C and D lie on the same side of AB.
👉 Wrong: Any chord can be used to claim that an angle is 90°.
Correct: The chord must be a diameter for the angle-in-a-semicircle theorem.
👉 Wrong: From ∠AOB = 2∠ACB, writing ∠ACB = 2∠AOB.
Correct: ∠ACB = 1/2 ∠AOB.
👉 Wrong: Always using the minor central angle for a major arc.
Correct: Use the central angle corresponding to the same arc as the inscribed angle. For a major arc, this may be a reflex central angle.
📚 Important points to remember
👉 The central angle standing on an arc is twice the inscribed angle standing on the same arc.
👉 All inscribed angles in the same segment are equal.
👉 If a segment subtends equal angles at two points on the same side, the four related points are concyclic.
👉 A diameter subtends an angle of 180° at the centre.
👉 The angle in a semicircle is 90°.
👉 A triangle formed by the endpoints of a diameter and any other point on the circle is right-angled; the diameter is its hypotenuse.
👉 To find an inscribed angle from the corresponding central angle, divide by 2.
👉 To find the corresponding central angle from an inscribed angle, multiply by 2.
👉 Before using a theorem, first identify clearly which chord or arc subtends which angle.