Arc - Properties of a Circle and Theoem, Class 9, Mathematics PDF Download

ARC PROPERTIES OF A CIRCLE

Theorem-10. In equal circles (or in the same circle), if two arcs subtend equal angles at the centre, they are equal.
 Given :Two equal circles C₁ and C₂ with O and O' as their centres respectively. AB subtends ∠AOB

and CD subtends ∠CO'D such that ∠AOB = ∠CO'D.
To prove : AB = CD

Proof :

So, AB = CD

Hence Proved

Converse of above theorem : In equal circles (or in the same circle), if two arcs are equal, they subtend equal angles at the centre.
 Given : Two equal circles C₁ and C₂ with O and O' as their respective centres such that AB = CD .

To prove : ∠AOB = ∠CO'D

Proof :

Hence Proved.

Theorem-11. In equal circles (or in the same circle), if two chords are equal, they cut off equal arcs.
 Given :Two equal circles C₁ and C₂ with centres O And, chord AB = chord CD.

To prove : AB = CD
Proof :

Hence Proved.

Converse of above theorem : In equal circles (or in the same circle), if two arcs are equal, then their chords are equal.
 Given : Two equal circles C₁ and C₂ with centres O
and, O' respectively and AB = CD .
To prove : 

Chord AB = Chord CD. 

Construction : Join OA, OB, O'C and O'D.
Proof :

Hence Porved.

Ex.25 In the given diagram, O is the centre of the circle and chord AB = chord BC.

(i) What is the relation between arc AB and arc BC ?
 (ii) What is the relation between ∠AOB and ∠BOC ?

Sol. (i) Since equal chords in a circle cut equal arcs, so chord AB = chord BC  arc AB = arc BC.
(ii) Equal arcs in a circle make equal angles at the centre.
∴ chord AB = chord BC ⇒ arc AB = arc BC. ⇒ ∠AOB = ∠BOC.

Ex.26 In the adjoining figure A, D, B, C are four points on the circumference of a circle with centre O.

(i)  [∵ Angle at the centre is double the ∠ at a point on the circumference]

(ii) ∠CAB = ∠COB = × ∠BOC =× 54° = 27°.

(iii) ∠ADB = reflex ∠AOB = (360° – 108°) = × 252° = 126°.

CHORD PROPERTIES OF A CIRCLE 

Theorem-1 : Equal chords of a circle subtend equal angles at the centre.
Given : A circle with centre O in which chord PQ = chords RS. P

To prove : ∠POQ = ∠ROS
Proof : In ΔPOQ and ΔROS,

Hence proved

Converse of above theorem : If the angles sutended by the chords at the centre (of a circle) are equal, then the chords are equal.
Given : A circle with centre O. Chord PQ and RS subtend equal angles at the centre of the circle.
i.e., ∠POQ = ∠ROS P

To prove : Chord PQ = Chord RS O
Proof : In ΔPOQ and ΔROS,

Hence proved

Theorem 2 : The perpendicular from the centre of a circle to chord bisects the chord.
Given : AB is a chord of a circle with centre O.
To prove : LA = LB.

Construction : Join OA and OB. L
Proof :

Hence proved

Converse of above theorem : The straight line drawn from the centre of a circle to bisect a chord, is perpendicular to the chord.
Given : AB is chord of a circle with centre O and OL bisects AB.
To prove: OL ⊥ AB.

Construction : Join OA and OB. L
Proof :

Prove that one and only one circle, passing through three non-collinear points.
 Given :Three non-collinear points A, B, C.
To prove : One and only one circle can be drawn, passing through A, B, and C.

Construction : Join AB and BC. Draw the perpendicular bisectors of AB O and BC. Let these perpendicular bisector intersect meeting at a point O.
Proof :

Hence, one and only circle can be drawn through three non-collinear points A, B and C.