Question

In $△ABC$, angle $ABC$ is equal to twice the angle $ACB$, and bisector of angle $ABC$ meets the opposite side at point $P$, show that:
(i) $CB:BA=CP:PA$
(ii) $AB\times BC=BP\times CA$

Answer 1

In $△ABC$

$\angle ABC=2\angle ACB$ ( given)

Let $\angle ACB=x$

$BP$ is bisector of $\angle ABC$

$\therefore \angle ABP=\angle PBC=x$

By angle bisector theorem

the bisector of an divides the side opposite to it in ratio of other two sides

Hence, $\frac{AB}{BC}=\frac{CP}{PA}$

$\therefore \frac{CB}{BA}=\frac{CP}{PA}$

Consider $△ABC$ and $△APB$

$\angle ABC=\angle APB$ (exterior angle properly)

given that, $\angle BCP=\angle ABP$

$\therefore ABC\approx △APB$ ($AA$ criteria)

$\therefore \frac{AB}{BP}=\frac{CA}{CB}$ (corresponding sides of similar triangle are propositional)

$\therefore AB\times BC=BP\times CA$

Hence proved