In figure,\(\angle{PQR}\) = \(\angle{PRQ}\), then prove that\(\angle{PQS}\) = \(\angle{PRT}\) .

By Linear pair axiom,
\(\angle{PQS}\) + \(\angle{PQR}\) = \(180^\circ\)...(i)

Similarly, again by Linear pair axiom,
\(\angle{PRQ}\) + \(\angle{PRT}\) = \(180^\circ\) ...(ii)

Thus by Eq. (i) and (ii),we get,
\(\angle{PQS}\) + \(\angle{PQR}\) = \(\angle{PRQ}\) + \(\angle{PRT}\)
As it is given, \(\angle{PQR}\) = \(\angle{PRQ}\)
\(\therefore \) \(\angle{PQS}\) + \(\angle{PRQ}\) = \(\angle{PRQ}\) + \(\angle{PRT}\)
By cancelling equal terms,
we get, \(\angle{PQS}\) = \(\angle{PRT}\)
Hence, proved.