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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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