XY ​ is the perpendicular bisector of PQ . Then ∆POX ≅ ∆QOX by ​

Question

\(\overline{XY}\)is the perpendicular bisector of \(\overline{PQ}\) . Then ∆POX ≅ ∆QOX by

A) S.A.S 

B) A.S.A 

C) S.S.S 

D) A.A.A.

Answer 1

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)

Answer 2

Correct option is  A) S.A.S