proof of bisectors theorem

Consider sines law in triangles $\mathrm{△}APB$ and $\mathrm{△}APC$.

On $\mathrm{△}APB$ we have

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

and on $\mathrm{△}APC$ 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, $\mathrm{\angle }APB+\mathrm{\angle }CPA={180}^{\circ }$, 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 $AP$ is a bisector, $\mathrm{\angle }BAP=\mathrm{\angle }PAC$ and we can cancel further to obtain the bisector theorem.