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In two concentric circles, prove that all chords of the outer circle which touch the inner circle, are of equal length.

 

Answers (1)

Given \rightarrow

C_{out}\rightarrow outer \; circle

C_{in}\rightarrow inner \; circle

AB and CD chords of outer circle tangent to the inner circle.

O\rightarrow common centre of C_{in} and C_{out} OE and OF are points of contact of C_{in}

To Prove \rightarrow AB=CD

Theorem \rightarrow Tangent at any point of the circle is perpendicular to the radius through the point of contact

According to the above theorem

\angle OEA=90^{o} and \angle OFC=90^{o}\; \; \; \; \; \; \; -----(1)

Now in \Delta AOE\; and\; \Delta COF

OA=OC    ( Radius of C_{out})

OE = OF     ( Radius of C_{in})

\angle OEA=\angle OFC=90^{o}          (from (i))

Now using R.H.S congruency rule

     \Delta AOE\cong \Delta COF

This means \Rightarrow\Rightarrow EB=DF\; \; \; \; \; -----(iii)

Similarly \Delta BOE and \Delta DOF are congruent

AE=FC\; \; \; \; \; \; \; --------(ii)

Adding (i) and (iii)

AE+EB=FC+DF

\Rightarrow AB=CD

Hence Proved

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Safeer PP

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