Equal Chords are Equidistant from the Center of Circle


 
 
Concept Explanation
 

Equal Chords are Equidistant from the Center of Circle

Theorem 1   Equal chords of a circle are equidistant from the centre

Given       Two  chords AB and CD of a circle C (O,r) such that AB = CD and OL large perp AB and OM large perp CD.

To prove     Chords AB and CD are equidistant from the centre O i.e, OL = OM.

CONSTRUCTION    Join OA and OC.

PROOF  Since the perpendicular from the centre to a chord bisects the chord

Therefore,

            large OL perp AB Rightarrow AL =frac{1}{2} AB              large ...(i)

and,     large OM perp CD Rightarrow CM = frac{1}{2} CD        large ...(ii)

But,    large AB = CD

large Rightarrow frac{1}{2} AB = frac{1}{2}CD

large Rightarrow AL = CM               large [using (i) and (ii)] ...(iii)

Now, in right triangles OAL and OCM, we have

       large OA = OC               [ Equal to radius of the circle ]

      large AL = CM              [ From equation (iii) ]

and, large angle ALO = angle CMO  [ Each equal to large 90^{circ}]

So, by RHS criterion of convergence, we have

        large Delta OAL cong Delta OCM

large Rightarrow OL = OM

Hence, equal chords of a circle are equidistant from the centre.

Theorem 2   Chords of a circle which are equidistant from the centre are equal.

Given    Two chords AB and CD of a circle C(O,r) which are equidistant from its centre i.e., OL = OM, where OL large perp AB and OM large perp CD.

TO PROVE   Chords are equal i.e AB = CD

CONSTRUCTION  Join OA and OC

PROOF    Since the perpendicular from the centre of a circle to a chord bisects the chord.

Therefore,

          large OL perp AB

large Rightarrow AL = BL

large Rightarrow AL = frac{1}{2} AB                        ...(i)

and,    large OM perp CD

large Rightarrow CM = DM

large Rightarrow CM = frac{1}{2}CD                     ...(ii)

In, triangles OAL and OCM , we have

    large OA = OC                  [ Each equal to radius of the given circle ]

large angle OLA = angle OMC          [ Each equal to large 90^{circ}]

and,  large OL = OM                            [ Given ]

So, by RHS< criterion of convergence, we have

        large Delta OAL cong Delta OCM

large Rightarrow AL = CM

large Rightarrow frac{1}{2}AB = frac{1}{2}CD

large Rightarrow AB = CD                      [ Using (i) and (ii) }

Hence, the chords of a circle which are equidistant from the centre are equal.

Illustration: 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

AB and CD are equal chords of a circle with O. The chords intersect each other at E within the circle.. If large angleAEO = large 30^0, What is the measure of large angleEON ?

 

Right Option : C
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Explanation
Question : 2
In the given, O is the centre of the circle, AB = BC = 4 units and ON = 4.5 units. The value of OM + MB is equal to

 

Right Option : B
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Explanation
Question : 3

If arc;RQ cong arc; PS and angle ROQ=60^o, then  angle POS is equal to

 

Right Option : C
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Explanation
 
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