iit jee geometry experts

its a JEE 2008 quesn !!!
oofff............JEE 2008 was really very hard man!!!:(

(PS) (ST) = (QS) (SR)

now use AM > GM

1/2 ( 1/PS + 1/ST ) > 1/ √(PS) (ST)

1/PS + 1/ST > 2/√(QS) (SR)

now if x+ y = k (a constant ) , then xy is max. when x = y = k/2

thus (QS ) (SR) < 1/4 (QR)²

1/(QS ) (SR) > 4/(QR)²

thus , 1/PS + 1/PT > 2√ 4/(QR)² = 4/QR

so B and D r correct..

apply, similarity in triangle PSQ and TSR.
angle PQS = angle STQ
and angle PSQ = angle TSR.

thus,
triangle PSQ and TSR are similar
thus, PSSR = QSST.