Q5 of 31 Page 209

Two chords AB and AC of the larger of two concentric circles touch the other circle at points P and Q respectively. Let us prove that, PQ = 1/2 BC.

Formula used.


⇒ Mid-point theorem.


Line joining mid-point of 2 sides of triangle is half of 3rd side of triangle


⇒ Perpendicular to chord divides the chord in 2 equal parts


Solution



As AB and AC are tangents for smaller circle


And Chords for bigger circle


The perpendicular of tangents passes through centre


And radius of smaller circle act as perpendicular to chord


Perpendicular of chord divides it into equal parts


∴ P and Q are mid-points of AB and AC


Join BC to form Δ ABC


As P and Q are mid-points of AB and AC


By mid-point theorem


Line joining mid-point of 2 sides of triangle is half of 3rd side


Hence;


PQ = 1/2 BC

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