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Question
PQ and QR are two equal chords of a circle. A diameter of the circle is drawn through Q . Prove that the diameter bisects ∠ PQR.
Solution
Let QT be the diameter of ∠ PQR
Since , PQ = QR
∴ OM = ON
In Δ OMQ and Δ ONQ
OM = ON (equal chords are equidistant from the centre)
∠ OMQ = ∠ ONQ (90 ° each)
OQ = OQ (common)
Δ OMQ ≅ Δ ONQ (RHS)
∴ ∠ OQM = ∠ OQN (CPCT)
Thus QT i.e. diameter of the circle visects ∠ PQR.
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