In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
![](/media/uploads/2021/12/In-the-given-figure-sides-AB-and-AC-of-ΔABC-are-extended-to-points-P-and-Q-respectively..png)
Solution:
In the given figure,
$\angle \mathrm{ABC}+\angle \mathrm{PBC}=180^{\circ}$ (Linear pair)
$\Rightarrow \angle \mathrm{ABC}=180^{\circ}-\angle \mathrm{PBC} \ldots(1)$
Also,
$\angle \mathrm{ACB}+\angle \mathrm{QCB}=180^{\circ}$
$\angle A C B=180^{\circ}-\angle Q C B \ldots(2)$
$\mathrm{As} \angle \mathrm{PBC}<\angle \mathrm{QCB}$ $\Rightarrow 180^{\circ}-\angle \mathrm{PBC}>180^{\circ}-\angle Q C B$
$\Rightarrow \angle A B C>\angle A C B$ [From equations (1) and (2)]
$\Rightarrow A C>A B$ (Side opposite to the larger angle is larger.)
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