In Figure, \(PS\) is the bisector of \(\angle P\) and \(PQ=PR\) .Then \(\Delta PRS\) and \(\Delta PQS\) are congruent by the criterion

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D

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In \(\Delta PRS\) and \(\Delta PQS\)

\(PQ=PR\) [Given]

\(\angle QPS = \angle RPS\) [Given]

\(PS =PS\) [Common]

Hence \(\Delta PRS \cong \Delta PQS\) [SAS criteria]

Also, In \(\Delta PRS\) and \(\Delta PQS\)

\(\angle PQS=\angle PRS\) (Angles opposite to equal sides)

\(PQ=PR\) (Given)

\(\angle QPS=\angle RPS\) (Given)

Therefore, \(\Delta PRS \cong \Delta PQS\) (By ASA congruency criterion)

Hence, option (D) is correct.

\(PQ=PR\) [Given]

\(\angle QPS = \angle RPS\) [Given]

\(PS =PS\) [Common]

Hence \(\Delta PRS \cong \Delta PQS\) [SAS criteria]

Also, In \(\Delta PRS\) and \(\Delta PQS\)

\(\angle PQS=\angle PRS\) (Angles opposite to equal sides)

\(PQ=PR\) (Given)

\(\angle QPS=\angle RPS\) (Given)

Therefore, \(\Delta PRS \cong \Delta PQS\) (By ASA congruency criterion)

Hence, option (D) is correct.

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