In figure, PR > PQ and PS bisects \(\angle{QPR}\). Prove that \(\angle{PSR}\) > \(\angle{PSQ}\).

In \(\triangle{PQR}\), we have,
PR > PQ ...(given)
\(\therefore \) \(\angle{PQR}\) > \(\angle{PRQ}\) ...(i)
(\(\because \) angle opposite to longer side is greater)
We can also say,
\(\angle{a}\) = \(\angle{b}\) ...(ii)
(\(\because \) PS bisects \(\angle{QPR}\))



On adding Equation (i) and (ii), we get,

\(\angle{PQR}\) + \(\angle{a}\)> \(\angle{PRQ}\) + \(\angle{b}\) ...(iii)
Also, \(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(180^\circ\) ...(iv)
and \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\) = \(180^\circ\) ...(v)
(\(\because \) the sums of all angles of a triangle are \(180^\circ\))

From Equation (iv) and (v), we get,

\(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\)
\(\because \) \(\angle{PQS}\) = \(\angle{PQR}\) and \(\angle{PRS}\) = \(\angle{PRQ}\),
we get,

\(\angle{PQR}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRQ}\) + \(\angle{PSR}\) + \(\angle{b}\)

Rearranging the terms, we get,

\(\angle{PQR}\) + \(\angle{a}\) + \(\angle{PSQ}\) = \(\angle{PRQ}\) + \(\angle{b}\) + \(\angle{PSR}\) ...(vi)

From Equation (iii) and (vi), we get,

\(\angle{PSQ}\) < \(\angle{PSR}\) ...
(\(\because \) side opposite to greater angle is longer)
Hence, proved.