In the figure, PQ = PR, show that PS > PQ.

In ΔPQR, PQ = PR
      ⇒ ∠PQR = ∠PRQ ..(i)
Now in ΔPSQ, ∠PQR is the exterior angle
      ⇒ ∠PQR = ∠PSQ + ∠SPQ
      ⇒ ∠PQR > ∠PSQ
      ⇒ ∠PRQ > ∠PSQ ...... (∠PQR = ∠PRQ)
          ⇒ PS > PR ...... (side opposite to greater angle)
         .^.. PS > PQ ...... (PR = PQ)