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Answer :
By linear pair axiom, we say that,
\(\angle{ACB}\) + \(\angle{QCB}\) = \(180^\circ\) ...(i)
\(\angle{ABC}\) + \(\angle{PBC}\) = \(180^\circ\) ...(ii)
From Equation (i) and (ii), we get,
\(\angle{ACB}\) + \(\angle{QCB}\) = \(\angle{ABC}\) + \(\angle{PBC}\)
But \(\angle{PBC}\) < \(\angle{QCB}\) ...(given)
\(\therefore \) \(\angle{ABC}\) > \(\angle{ACB}\)
\(\therefore \) AC > AB.
Hence, proved.