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Question
Given: PQ is perpendicular bisector of side AB of the triangle ABC.
Prove: Q is equidistant from A and B.
Solution
Construction: Join AQ
Proof: In ΔAQP and ΔBQP
AP = BP ...(Given)
∠QPA = ∠QPB ...(Each = 90°)
PQ = PQ ...(Common)
By side – Angle – side criterion of congruence, we have
ΔAQP ≅ ΔBQP ...(SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ ...(C.P.C.T.)
Hence Q is equidistant from A and B.
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