In the adjoining figure, centre of two circles is O. Chord AB of bigge

1.	In the adjoining 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.Given: Two concentric circles having centre O.To prove: AP = BQConstruction: Draw seg OM ⊥ chord AB, A-M-B
Answer» Proof: For smaller circle, seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B] ∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] For bigger circle, seg OM ⊥ chord AB [Construction] ∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] ∴ AP + PM = MQ + QB [A-P-M, M-Q-B] ∴ AP + MQ = MQ + QB [From (i)] ∴ AP = BQ

In the adjoining 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.Given: Two concentric circles having centre O.To prove: AP = BQConstruction: Draw seg OM ⊥ chord AB, A-M-B

Answer»

Proof:

For smaller circle,

seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B]

∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]

For bigger circle,

seg OM ⊥ chord AB [Construction]

∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]

∴ AP + PM = MQ + QB [A-P-M, M-Q-B]

∴ AP + MQ = MQ + QB [From (i)]

∴ AP = BQ

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