Question

In the given 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

In the given figure,
$\angle ABC+\angle PBC={180}^{\circ }$ (Linear pair)
$\therefore \angle ABC={180}^{\circ }-\angle PBC$ .........(1)
Also,
$\angle ACB+\angle QCB={180}^{\circ }$
$\therefore \angle ACB={180}^{\circ }-\angle QCB$ ........(2)
It is given that$\angle PBC<\angle QCB$
$\Rightarrow {180}^{\circ }-\angle PBC$ > ${180}^{\circ }-\angle QCB$
$\Rightarrow \angle ABC$ > $\angle ACB$ [From (1) and (2)]
$\therefore$ AC>AB.(In any triangle, the side opposite to the larger angle is longer)