Chords equidistant from the centre of a circle are equal in length.
We now take an example to illustrate the use of the above results:

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Solution

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Given, $d_{1} = d_{2}$
As perpendicular from center bisects the chord, $\Rightarrow B M - \frac{l_{1}}{2}, N D = \frac{l_{2}}{2}$
In $Δ O M B$ , by Pythagoras, $O B^{2} = d_{1}^{2} + (\frac{l_{1}}{2})^{2}$
Also in $Δ O N D, O D^{2} = d_{2}^{2} + (\frac{l_{2}}{2})^{2}$
As OB=OD = radius $\Rightarrow (\frac{l_{1}}{2})^{2} = (\frac{l_{2}}{2})^{2} \Rightarrow l_{1} = l_{2}$

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