Correct option is (C) right
In triangles \(\triangle PAQ\;and\;\triangle PBQ,\)
PA = PB, (Given)
AQ = BQ, (Given)
PQ = PQ (Common side)
\(\therefore\) \(\triangle PAQ\cong\triangle PBQ\) (By SSS congruence criteria))
\(\therefore\) \(\angle APQ=\angle BPQ\) (By corresponding property of congruence triangles)
\(\Rightarrow\) \(\angle APO=\angle BPO\) _______(1)
Now, in triangles \(\triangle APO\;and\;\triangle BPO,\)
AP = BP (Given)
\(\angle APO=\angle BPO\) (From (1))
OP = OP (Common side)
\(\therefore\) \(\triangle APO\cong\triangle BPO\) (By SAS congruence criteria)
\(\therefore\) \(\angle AOP=\angle BOP\) _______(2) (By corresponding property of congruence triangles)
Since, \(\angle AOP\;and\;\angle BOP\) forms a linear pair.
\(\therefore\) \(\angle AOP+\angle BOP\) = \(180^\circ\)
\(\Rightarrow\) \(2\angle AOP\) = \(180^\circ\) \((\because\) \(\angle AOP=\angle BOP\) (From (2)))
\(\Rightarrow\) \(\angle AOP\) = \(\frac{180^\circ}2=90^\circ\)
\(\Rightarrow\) \(\angle POA\) = \(90^\circ\)
Hence, \(\angle POA\) is a right angle.
