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prove that equal chords of circle are equidistant from the centre

 consider a circle of centre O.It has AB and CD two chords of equal length.now draw a perpendicular from centre to AB and CD intersecting them at P and Q.so we have to prove that OP=OQ.join OB and OD.

we know that AB=CD(given)

BP=AB/2(perpendicular drawn from a centre to a chord bisect the chord)

similarly  DQ=DC/2

therefore BP=DQ

consider triangleOBP and triangle ODQ

OB =OD(radius of the same circle)

BP=DQ(proved)

angleOPB=angleODQ=90(given)

triangleOBP is congruent to triangleODQ

this implies that OP=OQ(By CPCT)

Therefore equal chords of a circle are equidistant from the centre

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Given, two chords AB and CD  of circle with centre O, which are equidistant from its centre i.e., OL = OM, where OL ⊥ AB and OM ⊥ CD.

To Prove: AB = CD

Construction: Join OA and OC.

Proof:

OL ⊥ AB

⇒AL = BL [ Since, the ⊥cular from the centre of a circle to a chord bisects the chord ]

and, OM ⊥ CD

⇒ CM = DM [ Since, the ⊥cular from the centre of a circle to a chord bisects the chord ]

In ∆OAL and ∆OCM, we have

OA = OC  [ Radii of given circle ]

∠OLA = ∠OMC  [ Each equal to 90° ]

and OL = OM  [ Given ]

∴ ∆OAL ≅ ∆OCM [ RHS congruency ]

⇒ AL = CM

⇒ AB = CD

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Given,two chords AB and CD of circle with centre O, which are equidistant from its centre i.e., OL = OM, where OL ⊥ AB and OM ⊥ CD.

To Prove:AB = CD

Construction:Join OA and OC.

Proof:

OL⊥ AB

⇒AL = BL [ Since, the⊥cular from the centre of a circle to a chord bisects the chord ]

and, OM⊥ CD

⇒ CM = DM [ Since, the⊥cular from the centre of a circle to a chord bisects the chord ]

In ∆OAL and ∆OCM, we have

OA = OC [ Radii of given circle ]

OLA = OMC [ Each equal to 90 ]

and OL = OM [ Given ]

∴ ∆OAL ≅ ∆OCM [ RHS congruency ]

⇒ AL = CM

⇒ AB = CD

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