# 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 $AP$ is a bisector, $\angle BAP=\angle PAC$ and we can cancel further to obtain the bisector theorem.

Title proof of bisectors theorem ProofOfBisectorsTheorem1 2013-03-22 14:49:28 2013-03-22 14:49:28 drini (3) drini (3) 5 drini (3) Proof msc 51A05