Class 10 Mathematics Chapter 20 Angle Properties of A Circle: ICSE Study Material
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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 for Class 10 Mathematics Chapter 20 Angle Properties of A Circle
ICSE Book Class 10 Mathematics Chapter 20 Angle Properties of A Circle
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