Question

In the adjoining figure, sides $AB$ and $AC$ of $△ABC$ are extended to points $P$ and $Q$ respectively.Also, $\angle PBC<\angle QCB$. Show that $AC>AB.$

Answer 1

REF.Image

Given

$\angle PBC<\angle QCB$

$\angle PBC$ is the external angle of $\Delta ABC$

so, $\angle PBC=\angle A+\angle ACB$

Now,

$\angle PBC<\angle QCB$

$\Rightarrow \angle A+\angle ACB<\angle A+\angle ABC$

$\Rightarrow \angle ACB<\angle ABC$

$\Rightarrow AB<AC(\because$ Side opposite to the greater angle is longer)

Thus

AC > AB

Hence Proved