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ICSE Class 10 Mathematics Chapter 20 Angle Properties of A Circle Digital Edition
For Class 10 Mathematics, this chapter in ICSE Class 10 Maths Chapter 20 Angle Properties of A Circle provides a detailed overview of important concepts. We highly recommend using this text alongside the ICSE Solutions for Class 10 Mathematics to learn the exercise questions provided at the end of the chapter.
Chapter 20 Angle Properties of A Circle ICSE Book Class Class 10 PDF (2026-27)
Chapter 20
Angle Properties Of A Circle
Points To Remember
1. Some Important Theorems
Theorem 1. The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given. A circle with centre O and an arc AB subtends \(\angle AOB\) at the centre and \(\angle ACB\) at any point C on the remaining part of the circle.
To prove. \(\angle AOB = 2 \angle ACB\).
Construction. Join CO and produce it to some point D.
Proof.
| Statement | Reason |
|---|---|
| 1. In \(\triangle AOC\), \(OA = OC\) \(\Rightarrow \angle OAC = \angle OCA\) ....(I) | Radii of the same circle. Angles opposite to equal sides of a triangle are equal. |
| 2. \(\angle AOD = \angle OCA + \angle OCA = \angle OCA + \angle OCA = 2 \angle OCA\) ....(II) | Ext. angle of a triangle = Sum of its int. opp. angles. Using (I). |
| 3. Similarly, \(\angle BOD = 2 \angle OCB\) ....(III) | |
| 4. In figure (i), \(\angle AOD + \angle BOD = 2 \angle OCA + 2 \angle OCB = 2 (\angle OCA + \angle OCB) = 2 \angle ACB\) \(\therefore \angle AOB = 2 \angle ACB\). | Adding corresponding sides of (II) and (III). |
Teacher's Note
When you look at a circular clock, the angle at the center (formed by hour and minute hands) is always twice the angle you would measure from any point on the clock's rim - this is the essence of this theorem in action.
In Figure (iii), \(\angle AOD + \angle BOD = 2 \angle OCA + 2 \angle OCB = 2 (\angle OCA + \angle OCB) = 2 \angle ACB\). \(\therefore\) Reflex \(\angle AOB = 2 \angle ACB\). Adding the corresponding sides of (II) and (III).
In Figure (ii), \(\angle BOD - \angle AOD = 2 \angle OCB - 2 \angle OCA = 2 (\angle OCB - \angle OCA) = 2 \angle ACB\). \(\therefore \angle AOB = 2 \angle ACB\). Subtracting the corresponding sides of (III) and (II).
Hence, \(\angle AOB = 2 \angle ACB\).
Theorem 2. Angles in the same segment of a circle are equal.
Given. A circle with centre O and two angles \(\angle ACB\) and \(\angle ADB\) in the same segment of the circle.
To prove. \(\angle ACB = \angle ADB\).
Construction. Join OA and OB.
Proof.
| Statement | Reason |
|---|---|
| In Fig. (I): 1. Arc AB subtends \(\angle AOB\) at the centre and \(\angle ACB\) at a point C of the remaining part of the circle. \(\therefore \angle AOB = 2 \angle ACB\) ....(I) | Angle at the centre is double the angle at any point on remaining part of the circle. |
| 2. Arc AB subtends \(\angle AOB\) at the centre and \(\angle ADB\) at a point D on the remaining part of the circle. \(\therefore \angle AOB = 2 \angle ADB\) ....(II) | Same as above. |
| 3. 2 \(\angle ACB = 2 \angle ADB\) \(\therefore \angle ACB = \angle ADB\) | From (I) and (II). |
| 4. Similarly, in Fig. (II): \(\angle ACB = \angle ADB = \frac{1}{2}\) reflex \(\angle AOB\) \(\therefore \angle ACB = \angle ADB\). |
Hence, the angles in the same segment of a circle are equal.
Theorem 3. The angle in a semi-circle is a right angle.
Given. A semi-circle ACB of a circle with centre O.
To prove. \(\angle ACB = 90°\).
Proof.
| Statement | Reason |
|---|---|
| 1. Arc AB subtends \(\angle AOB\) at the centre and \(\angle ACB\) at a point C on the remaining part of the circle. \(\therefore \angle AOB = 2 \angle ACB\) \(\Rightarrow \angle ACB = \frac{1}{2} \angle AOB\) ....(I) | Angle at the centre is double the angle at any point on remaining part of the circle. |
| 2. \(\angle AOB = 180°\) ....(II) | AOB is a straight line. |
| 3. \(\angle ACB = \left(\frac{1}{2} \times 180°\right) = 90°\) | From (I) and (II). |
Hence, the angle in a semi-circle is right angle.
Theorem 4 (Converse of Theorem 3). If an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a semi-circle.
Given. A circle with centre O and an arc AB subtending \(\angle ACB\) at a point C on the remaining part of the circle such that \(\angle ACB = 90°\).
To prove. Arc AB is a semi-circle.
Construction. Join OA and OB.
Proof.
| Statement | Reason |
|---|---|
| 1. Arc AB subtends \(\angle AOB\) at the centre and \(\angle ACB\) at a point C on the remaining part of the circle. \(\therefore \angle AOB = 2 \angle ACB\) ....(I) | Angle at the centre is double the angle at a point on the remaining of the circle. |
| 2. \(\angle ACB = 90°\) ....(II) | Given. |
| 3. \(\therefore \angle AOB = (2 \times 90°) = 180°\) \(\Rightarrow AOB\) is a straight line. \(\Rightarrow AOB\) is a diameter \(\Rightarrow\) Arc AB is a semi-circle. | From (I) and (II). Chord AB passes through centre O. |
Hence, arc AB is a semi-circle.
Teacher's Note
When you draw a triangle with any point on a circle and two endpoints of a diameter as vertices, that triangle will always have a right angle - this principle is used in construction and engineering to ensure perpendicular lines.
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ICSE Book Class 10 Mathematics Chapter 20 Angle Properties of A Circle
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