Question

In 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.

Answer 1

$\angle PBC$ is the exterior angle of $△ABC$.

Since exterior angle is sum of interior opposite angles.

So,

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

similarly,

$\angle QCB$ is the exterior angle of $△ABC$

So,

$\angle QCB=\angle A+\angle ABC$

Now,

$\Rightarrow$ $\angle PBC<\angle QCB$

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

$\Rightarrow$ $\angle ABC<\angle ACB$

On writing the opposite sides of the angles, we get

$\Rightarrow$ $AB<AC$

Hence proved.