Question

In adjacent 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

Given side AB and AC of triangle ABC are extended to point P and Q respectively

$\angle PBC<\angle QCB$ ....Given ....(1)

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

So, $\angle PBC=\angle A+\angle ACB$ ....(2)

And the $\angle QCB$ is the external angle of $\Delta ABC$

So $\angle QCB=\angle A+\angle ABC$ ...(3)

From (1), (2), and (3), we get

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

$\Rightarrow \angle ACB<\angle ABC$

$\therefore AB<AC$