Given in ΔPSR,Q is a point on the side SR such that PQ = PR.
To prove PS > PQ
Proof in ΔPRQPQ=PR [given]
⇒∠R=∠RQR...(i)
[angles opposite to equal sides are equal]
But ∠PQR>∠S ....(ii)
[exterior angle of a triangle is greater than each of the opposite interior angle]
from Eqs. (i) and (ii) ∠R>∠S
⇒PS>PR [side opposite to greater angle is longer]
⇒PS>PQ[∵PQ=PR]