Theorem 10.6 : Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
Text Solution
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Since `OXbotAB`
Perpendicular from the center to the chord, bisects the chord,
`AX=BX=(AB)/2`
Similarly,
`CY=DY=(CD)/2`
`AB=CD` (given)
Thus,
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