In the figure,sides $AB$ and $AC$ of $\Delta ABC$ are extended to points $P$ and $Q$ respectively. Also,$\angle PBC < \angle QCB$. Show that $AC > AB$.
(N/A) We know that $\angle ABC + \angle PBC = 180^\circ$ [Linear pair axiom].
Similarly,$\angle ACB + \angle QCB = 180^\circ$ [Linear pair axiom].
Therefore,$\angle ABC + \angle PBC = \angle ACB + \angle QCB$.
Given that $\angle PBC < \angle QCB$.
Since the sum of the angles is constant $(180^\circ)$,if one angle in the sum is smaller,the other must be larger to maintain the equality.
Thus,$\angle ABC > \angle ACB$.
In any triangle,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ABC$ (which is $AC$) is greater than the side opposite to $\angle ACB$ (which is $AB$).
Hence,$AC > AB$.
Similarly,$\angle ACB + \angle QCB = 180^\circ$ [Linear pair axiom].
Therefore,$\angle ABC + \angle PBC = \angle ACB + \angle QCB$.
Given that $\angle PBC < \angle QCB$.
Since the sum of the angles is constant $(180^\circ)$,if one angle in the sum is smaller,the other must be larger to maintain the equality.
Thus,$\angle ABC > \angle ACB$.
In any triangle,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ABC$ (which is $AC$) is greater than the side opposite to $\angle ACB$ (which is $AB$).
Hence,$AC > AB$.
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