Question

In the 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

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

Multiply the equation by $-1.$

$\Rightarrow -\angle PBC>-\angle QCB$

Adding ${180}^{\circ }$ on both sides, we get

$\therefore {180}^{\circ }-\angle PBC>{180}^{\circ }-\angle QCB$

Angles on a straight line add to ${180}^o$

Sum of angles $\angle PBC$ and $\angle ABC$ is ${180}^o.$

Sum of angles $\angle QCB$ and $\angle ACB$ is ${180}^o.$

So,

$\Rightarrow \angle ABC>\angle ACB$

$\Rightarrow AC>AB$ (Side opposite to greater angle is greater)

Hence Proved