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Question
If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ arc PYB.
Solution
Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O.
To prove: arc PXA ≅ arc PYB
Construction: Join AP and BP.
Proof: In ΔAPM and ΔBPM,
AM = MB ...[∵ PM bisects AB]
∠PMA = ∠PMB ...[Each 90°, ∵ PM ⊥ AB]
PM = PM ...[Common]
∴ ΔAPM ≅ ΔBPM ...[By SAS congruency]
∴ PA = PB ...[By C.P.C.T.]
⇒ arc PXA ≅ arc PYB
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