# 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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