If a line intersects two concentric circles (circles with the same centre) with centre $ \mathrm{O} $ at $ \mathrm{A}, \mathrm{B}, \mathrm{C} $ and D, prove that $ \mathrm{AB}=\mathrm{CD} $.(see figure below)
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Given:

$A,B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^o$ and $\angle AOB = 60^o$.

$D$ is a point on the circle other than the arc $ABC$.

To do:

We have to find $\angle ADC$.

Solution:

Draw a line segment from $O$ to $AD$ such that $OP \perp AD$.

$OP \perp AD$

This implies,

$OP$ bisects $AD$

Therefore,

$AP = PD$..........(i)

$OP \perp BC$

This implies,

$OP$ bisects $BC$.

Therefore,

$BP = PC$............(ii)

Subtracting (ii) from (i), we get,

$AP-BP = PD-PC$

Therefore,

$AB = CD$

Hence proved.