In Fig. 7.48, sides $ \mathrm{AB} $ and $ \mathrm{AC} $ of $ \triangle \mathrm{ABC} $ are extended to points $ \mathrm{P} $ and $ \mathrm{Q} $ respectively. Also, $ \angle \mathrm{PBC}\mathrm{AB} $.
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Given:
 Sides $AB$ and $AC$ of $\triangle ABC$ are extended to points $P$ and $Q$ respectively. Also, $\angle PBC<\angle QCB$.

To do:

We have to show that $AC>AB$.

Solution:

We know that,

The sum of the measures of the angles in linear pairs is always $180^o$

This implies,

$\angle ABC+\angle PBC=180^o$

This implies,

$\angle ABC=180^o-\angle PBC$

In a similar way, we get,

$\angle ACB+\angle QCB=180^o$

This implies,

$\angle ABC=180^o-\angle QCB$

Given,

$\angle PBC<\angle QCB$,

This implies,

$\angle ABC>\angle ACB$

We know that,

The side opposite the larger angle is always larger,

Hence, $AC>AB$.