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Question
In the given figure, if seg PR ≅ seg PQ, show that seg PS > seg PQ.
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Solution
In ∆PQR,
seg PR ≅ seg PQ ...[Given]
∴ ∠PQR ≅ ∠PRQ ...(i) [Isosceles triangle theorem]
∠PRQ is the exterior angle of ∆PRS.
∴ ∠PRQ > ∠PSR ...(ii) [Property of exterior angle]
∴ ∠PQR > ∠PSR ...[From (i) and (ii)]
i.e. ∠Q > ∠S ...(iii)
In ∆PQS,
∠Q > ∠S ...[From (iii)]
∴ PS > PQ ...[Side opposite to greater angle is greater]
∴ seg PS > seg PQ
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