prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
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Step-by-step explanation:
Given: 2 concentric circles be C1 and C2 with centre O
AB be chord of the larger circle C1 which touches the smaller circle C2 at point P.
To Prove: Chord AB is bisected at point of contact i.e., AB=AP
Solution: since AB is a tangent to smaller circle C2
OP perpendicular to AB
Now,
AB is a chord of bigger circle C1 and OP perpendicular to AB
As perpendicular from the centre bisects the chord
Therefore OP would be bisector of chord AB
=>AP=BP
Hence Proved
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