# in two concentric circles, prove that all chords of the outer circle which touches the inner circle are of equal lengths.

Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively. To prove the given question, it is sufficient to prove
AB = CD. For this join OM, ON, OB and OD.

Let the radius of outer and inner circles be R and r respectively.

AB touches the inner circle at M.
∴ AB is a tangent to the inner circle
∴ OM⊥AB
⇒ BM = ½ of AB
⇒ AB = 2BM

Similarly ON⊥CD, and CD = 2DN Hence proved

• 84

let AB be the chord of outer circle and OM be the radius of the inner circle which touches the chord AB.

to prove ---

AM = BM

OM is perpendicular to AB.

in triangle OMB and triangle OMA,

OM = OM [ common ]

OB = OA  [ radius of outer circle. ]

<OMB = <OMA

tri. OMA ia congruent to tri. OMB .[ SAS rule ]

MB = MA [ cpct ]

hence proved...

HOPE THIS HELPS...

• -25
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