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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.