In figure, Q is a point on side SR of ∆PSR such that PQ = PR. Prove

1.	In figure, Q is a point on side SR of ∆PSR such that PQ = PR. Prove that PS > PQ.
Answer» In ∆PQR PQ = PR ∠PQR = ∠PRQ …(i) (angles opposite to equal sides of a triangle are equal) In ∆PSQ Ext. ∠PQR > ∠PSQ …(ii) From (i) and (ii), we get ∠PRQ > ∠PSQ ⇒ PS > PR (side opposite to greater angle is longer) ⇒ PS > PQ (∵ PQ = PR)

In figure, Q is a point on side SR of ∆PSR such that PQ = PR. Prove that PS > PQ.

Answer»

In ∆PQR

PQ = PR

∠PQR = ∠PRQ …(i) (angles opposite to equal sides of a triangle are equal)

In ∆PSQ

Ext. ∠PQR > ∠PSQ …(ii)

From (i) and (ii), we get

∠PRQ > ∠PSQ

⇒ PS > PR (side opposite to greater angle is longer)

⇒ PS > PQ (∵ PQ = PR)