Theorems On Arcs And Chords

Theorems on Arcs and Chords Theorem: If two arcs of a circle (or of congruent circles) are congruent, then the corresponding chords are equal. Given: Two congruent circles C (O, r ) and C (O ' , r ) in which arc AB = arc CD. To prove: AB=CD Construction: Join OA, OB, O 'C and O ' D

Proof: We suppose that arc AB and arc CD are minor arcs In OAB and O 'CD ,

OA = O 'C = r OB = O ' D = r and ∠AOB = ∠CO ' D Since arc AB = arc CD

Therefore ∠AOB = ∠CO ' D Hence Triangle OAB ≅ O 'CD

(ASA)

Therefore AB = CD

(cpctc)

Theorem: (Converse of Previous Theorem): If two chords of a circle (or of congruent circles) are equal, then corresponding arcs are congruent. Given: Two congruent circles C (O, r ) and C (O ' , r ) in which chord AB = chord CD To prove: arc AB = arc CD Construction: Suppose that AB and CD are not diameters. Join OA, OB, O 'C and O' D .

Proof: In Triangle OAB and Triangle O 'CD

OA = O 'C = r OB = O ' D = r and AB=CD

(Given)

Therefore Triangle OAB ≅ O 'CD

Hence, ∠AOB = ∠CO ' D This implies that m (arc AB) = m (arc CD) or arc AB ≅ arc CD Theorem: The perpendicular from the centre of a circle to a chord bisects the chord. Given: AB is a chord of a circle C (O, r ) and OM ⊥ AB . To prove: AM=BM Construction: Join OA and OB

Proof: In Triangle OAM andTriangle OBM , we have OA=OB

(Radii of the same circle)

OM=OM

(Common to both triangles)

∠OMA = ∠OMB

(Each = 90o as OM ⊥ AB )

Therefore Triangle OAM ≅ Triangle OBM (By RHS criterion of congruency)

Hence AM=BM (cpctc) Theorem: (Converse of Previous Theorem) The line joining the centre of a circle to the mid point of a chord is perpendicular to the chord. Given: AB is a chord of the circle with centre O and D is the mid point that is AD=DB To prove: OD ⊥ AB Construction: Join OA and OB Proof: In Triangle OAD and Triangle OBD , we have OA=OB

(Radii of the same circle are equal)

OD=OD

(Common to both triangles)

AD=DB

(Given)

Therefore Triangle OAD ≅ Triangle OBD (SSS criterion of congruence)

Hence, ∠ODA = ∠ODB

(cpctc)

But ∠ODA + ∠ODB = 180o

(Linear pair of angles)

or ∠ODA + ∠ODA = 180o

( Since ∠ODA = ∠ODB )

or 2∠ODA = 180o 180o or ∠ODA = = 90o 2 Now ∠ODA = ∠ODB = 90o

Therefore OD ⊥ AB Corollary Perpendicular bisectors of two chords of a circle intersect at its centre. Given: AB and CD are two chords of a circle C ( O, r ) and perpendicular bisectors MO ' and NO ' of AB and CD respectively meet at O ' . To prove: O ' coincides with centre O . Construction: Join OM and ON.

Proof: Since M is the mid point of the chord AB, therefore, OM must be perpendicular to AB. This means that OM and O ' M are the perpendicular bisector of AB. This is possible only if O ' N coincides with ON. Thus, we find that OM and O ' M are on one line and ON and O ' N are one line. Since two intersecting straight lines can intersect at only one point, therefore O ' must coincides with O. Hence, the perpendicular bisectors of AB and CD intersect at the centre of the circle. If they intersect at the centre, it proves that a perpendicular bisector of the chord of a circle passes through the centre of the circle. Theorem: There is one and only one circle passing through three non-collinear points. Given: Three non-collinear pints A, B and C. To prove: There is one and only one circle passing through the points A, B and C. Construction: Join AB and BC. Draw perpendicular bisectors PQ and RS of AB and BC respectively. Since A, B and C are not collinear, this implies that the perpendicular bisectors of PQ and RS are not parallel. Hence PQ and RS intersect at a point, say O. Join OA, OB and OC. Proof: Since O lies on the perpendicular bisector of AB, therefore OA=OB. Similarly, since O lies on the perpendicular bisector of BC, therefore OB=OC.

Therefore OA = OB = OC = r (suppose)

Thus, if we draw a circle with O as the centre and r as the radius, then the circle will pass through A, B and C. Uniqueness of this circle Since two lines cannot intersect in more than one point, therefore, the perpendicular bisector PQ and RS cannot intersect in more than one point. Hence, their point of intersection, O, is unique. Hence, there is one and only one circle passing through three non-collinear points A, B and C. Theorem 6: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centers) Case I: Same circle Given: AB and CD are equal chords of a circle C ( O, r ), OM ⊥ AB, ON ⊥ CD .

Proof: In right triangle Triangle OAM and Triangle OCN , we have

AM =

1 1 AB, CN = CD and AB=CD 2 2

This implies that AM=CN OA=OC

(Radii of the same circle)

∠OMA = ∠ONC

(each 90o )

Therefore Triangle OAM ≅ Triangle OCN (By RHS criterion of congruency) Therefore OM = ON Hence, equal chords of a circle are equidistant from the centre. Case II: Congruent Circle Given: AB and CD are equal chords of congruent circles C ( O, r ) and C ( O ' , r ) OM ⊥ AB, O ' N ⊥ D

To prove: OM = O ' N Construction: Join OA and O 'C

Proof: In right triangles Triangle OAM and O 'CN , we have

1 1 AB, CN = CD 2 2 Since AB = CD, AM =

Therefore AM = CN , ∠OMA = ∠O NC '

OA = O 'C

(each = 90o )

(radii of congruent circles)

Therefore OAM ≅ Triangle O 'CN (RHS criterion of congruency) Therefore Triangle OAM ≅ O 'CN Therefore OM = O ' N

(cpctc)

Hence, AB and CD are equidistant from O and O ' respectively. Theorem: (Converse of Previous theorem): (1) Chords of congruent circles which are equidistant from the corresponding centers are equal.

(2) Chords of a circle which are equidistant from the centre are equal. (1) Given: AB and CD are the chords of congruent circles C ( O, r ) and C ( O ' , r ) and OE ⊥ AB, O ' F ⊥ CD such that OE = O ' F To prove: OE = O ' F Construction: Join OA and O 'C Proof:

In a right triangles Triangle OAE and O 'CF

∠OEA = ∠O ' FC

(Each = 90o )

OA= O 'C

(Each= r )

AE=CF ( Since AE =

1 1 AB and CF = CD and AB=CD) 2 2

Therefore Triangle OAE ≅ Triangle O 'CF

(RHS criterion of congruency)

Hence OE = O ' F (cpctc) Thus, AB and CD are equidistant from O and O ' respectively (2) Given: A circle whose centre is O. AB and CD are two chords OP ⊥ AB and OQ ⊥ CD .

To prove: AB=CD Construction: Join OA and OC

In right triangles Triangle OAP and Triangle OCQ Hypotenuse OA= Hypotenuse OC (Radii of the same circle) OP = OQ (Given)

Therefore Triangle OAP ≅ Triangle OCQ Therefore AP = CQ But AP =

1 AB 2

(RHS criterion of congruency)

(cpctc) ( Since OP ⊥ AB and CQ ⊥ CD

1 and CQ = CD 2 This implies that AB=CD

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