The larger of two concentric circles has chords AB and AC that touch the smaller circle at points P and Q respectively. To prove: \(PQ = \frac{1}{2}BC\) Construction: Let O be the center of the circles. Join OA, OC, and OQ. Proof: In triangles ∆AOQ and ∆COQ: - \(AO = OC\) [radii of the same circle] - \(\angle OAQ = \angle OCQ\) [since AO = OC in ∆AOC] - OQ is the common side ∴ \(AQ = QC\) [corresponding sides] ∴ ∆AOQ ≅ ∆COQ ⇒ Q is the midpoint of AC Similarly, by joining OP and OB, it can be shown that P is the midpoint of AB. ∴ PQ is the line segment joining the midpoints of sides AB and AC of triangle ∆ABC. So, by the midpoint theorem: \[ PQ = \frac{1}{2}AB\quad \text{(Proved)} \]