Question

In the given figure, if seg PR ≅ seg PQ, show that seg PS > seg PQ. 

Answer 1

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