Q is a point on the side SR of a ΔPSR such that PQ = PR. Prove that PS > PQ.
Given: in ΔPSR, Q is a point on the side SR such that PQ = PR.
In ΔPRQ,
PR = PQ (given)
⇒ ∠PRQ = ∠PQR (opposite angles to equal sides are equal)
But ∠PQR > ∠PSR (exterior angle of a triangle is greater than each of opposite interior angle)
⇒ ∠PRQ > ∠PSR
⇒ PS > PR (opposite sides to greater angle is greater)
⇒ PS > PQ (as PR = PQ)