Correct option is (A) S.A.S.
In triangle \(\triangle POX\;and\;\triangle QOX,\)
PO = QO \((\because\) XY is bisector of PQ)
\(\angle POX=\angle QOX\) \((\because\) XY is perpendicular bisector of PQ)
OX = OX (Common side)
\(\therefore\) \(\triangle POX\cong\triangle QOX\) (By S.A.S. congruence rule)