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Question
Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB
Solution
Let C (O, r) and C(O', r) be two equal circles. clearly, C(O, r) ≅ C(O', r).
Since PQ is a common chord of two congruent circles.
Therefore,
arc PCQ = arc PDQ
⇒ ∠ QAP = ∠ QBP
Thus, in ΔQAB, we have
∠ QAP = ∠ QBP
⇒ QA = QB
Hence proved.
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