PQ and RQ are the chords of a circle equidistant from the centre. Prove that the diameter passing through Q bisects ∠PQR and ∠PSR.

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Solution

We know that the chords of a circle that are equidistant from the centre are equal.
∴ PQ = RQ
Now, in ΔPQS and ΔRQS, we have:
PQ = RQ (Equal chords)
∠QPS = ∠QRS = 90° (Angles in semicircle)
QS = QS (Common hypotenuse)
Thus, ΔPQS ≅ ΔRQS (RHS criterion)
i.e., ∠PQS = ∠ RQS and ∠PSQ = ∠ RSQ (CPCT)
Hence, the diameter QOS bisects ∠PQR and ∠PSR.

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