proof of bisectors theorem

Consider sines law in triangles $\triangle APB$ and $\triangle APC$ .

On $\triangle APB$ we have

$\frac{BP}{\sin BAP}=\frac{AB}{\sin APB}$

and on $\triangle APC$ we have

$\frac{PC}{\sin PAC}=\frac{AC}{\sin CPA}.$

Combining the two relation gives

$\frac{BP}{PC}=\frac{AB\sin BAP/\sin APB}{AC\sin PAC/\sin CPA}.$

However, $\angle APB+\angle CPA=180^{\circ}$ , and so $\sin APB=\sin CPA$ . Cancelling gives

$\frac{BP}{PC}=\frac{AB\sin BAP}{AC\sin PAC},$

which is the generalization of the theorem. When $A P$ is a bisector, $\angle BAP=\angle PAC$ and we can cancel further to obtain the bisector theorem.