In figure, sides AB and AC of are extended to points P and Q respectively. Also, \(\angle{PBC}\) < \(\angle{QCB}\). Show that AC > AB.

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.