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

1.	In the adjoining figure, if seg PR ≅ seg PQ, show that seg PS > seg PQ.
Answer» 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 APQS, ∠Q > ∠S [From (iii)] ∴ PS > PQ [Side opposite to greater angle is greater] ∴ seg PS > seg PQ

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

Answer»

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 APQS,

∠Q > ∠S [From (iii)]

∴ PS > PQ [Side opposite to greater angle is greater]

∴ seg PS > seg PQ