Notes On Arcs of a Circle - CBSE Class 9 Maths
A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent. Corresponding arcs of two equal chords of a circle are congruent. Congruent arcs of a circle subtend equal angles at the centre. Given: Two congruent arcs AB and CD. To prove: ∠ AOB = ∠COD Construction: Draw chords AB and CD. Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre. In the given figure, AB = CD (Chords corresponding to congruent arcs of a circle are equal) ∠AOB = ∠COD (Equal chords subtend equal angles at the centre) Hence, the theorem is proved. The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle. Given: Arc AB. Point C on the circle is outside AB. To prove: ∠AOB = 2 × ∠ACB Construction: Draw a line CO extended till point D. Proof: In ΔOAC in each of these figures, ∠AOD = ∠OAC + ∠OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles) OA = OC (Radii of same circle) Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal) ∠AOD = ∠OAC + ∠OCA ⇒∠AOD = 2 × ∠OCA Similarly, in ΔOBC, ∠BOD = 2 × ∠OCB ∠AOD = 2 × ∠OCA and ∠BOD = 2 × ∠OCB ⇒ ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB ∠AOD + ∠BOD = 2 × (∠OCA + ∠OCB) or ∠AOB = 2 × ∠ACB Hence, the theorem is proved. Angles subtended by an arc at all points within the same segment of the circle are equal. Given: An arc AB. Points C and D are on the circle in the same segment. To prove: ∠ACB = ∠ADB Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle: ∠AOB = 2 × ∠ACB Also, ∠AOB = 2 × ∠ADB ∴ ∠ACB = ∠ADB Hence, the theorem is proved. All angles formed in a semi circle are right angles. Given: A circle with centre O, and Q, P and R are three points on the circumference of the circle. Construction: Join the points Q, P and R to the points A and B. To prove: ∠AQB = ∠APB = ∠ARB Proof: ∠AQB = $\frac{\text{1}}{\text{2}}$ ∠AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.] ∠APB = $\frac{\text{1}}{\text{2}}$ ∠AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.] Similarly, ∠ARB = $\frac{\text{1}}{\text{2}}$ ∠AOB. ∴ ∠AQB = ∠APB = ∠ARB Hence, the theorem is proved.

#### Summary

A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent. Corresponding arcs of two equal chords of a circle are congruent. Congruent arcs of a circle subtend equal angles at the centre. Given: Two congruent arcs AB and CD. To prove: ∠ AOB = ∠COD Construction: Draw chords AB and CD. Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre. In the given figure, AB = CD (Chords corresponding to congruent arcs of a circle are equal) ∠AOB = ∠COD (Equal chords subtend equal angles at the centre) Hence, the theorem is proved. The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle. Given: Arc AB. Point C on the circle is outside AB. To prove: ∠AOB = 2 × ∠ACB Construction: Draw a line CO extended till point D. Proof: In ΔOAC in each of these figures, ∠AOD = ∠OAC + ∠OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles) OA = OC (Radii of same circle) Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal) ∠AOD = ∠OAC + ∠OCA ⇒∠AOD = 2 × ∠OCA Similarly, in ΔOBC, ∠BOD = 2 × ∠OCB ∠AOD = 2 × ∠OCA and ∠BOD = 2 × ∠OCB ⇒ ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB ∠AOD + ∠BOD = 2 × (∠OCA + ∠OCB) or ∠AOB = 2 × ∠ACB Hence, the theorem is proved. Angles subtended by an arc at all points within the same segment of the circle are equal. Given: An arc AB. Points C and D are on the circle in the same segment. To prove: ∠ACB = ∠ADB Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle: ∠AOB = 2 × ∠ACB Also, ∠AOB = 2 × ∠ADB ∴ ∠ACB = ∠ADB Hence, the theorem is proved. All angles formed in a semi circle are right angles. Given: A circle with centre O, and Q, P and R are three points on the circumference of the circle. Construction: Join the points Q, P and R to the points A and B. To prove: ∠AQB = ∠APB = ∠ARB Proof: ∠AQB = $\frac{\text{1}}{\text{2}}$ ∠AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.] ∠APB = $\frac{\text{1}}{\text{2}}$ ∠AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.] Similarly, ∠ARB = $\frac{\text{1}}{\text{2}}$ ∠AOB. ∴ ∠AQB = ∠APB = ∠ARB Hence, the theorem is proved.

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