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Question
In the given figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.
Solution
Given: Two concentric circles having centre O.
To prove: AP = BQ
Construction: Draw the line OM ⊥ chord AB such that, A-M-B
Proof:
With respect to the small circle, seg OM ⊥ chord PQ
∴ PM = MQ ...(i) ...(The perpendicular drawn from the center of the circle to the chord bisects the chord.)
Similarly, with respect to the great circle, seg OM ⊥ chord AB
∴ AM = MB ...(ii)
AM = AP + PM ...(A-P-M) ...(iii)
MB = MQ + QB ...(M-Q-B) ...(iv)
∴ AP + PM = MQ + QB ...[from (ii), (iii) and (iv)] ...(v)
∴ AP = BQ ...[from (i) and (v)]
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