# proof of bisectors theorem

Notice that the triangles $\triangle BAP$ and $\triangle CAP$ have the same common height $h$, and if $(BAP)$ and $(CAP)$ denote their respective areas, we have

 $\frac{(BAP)}{(CAP)}=\frac{BP\cdot h/2}{PC\cdot h/2}=\frac{BP}{PC}.$

On the other hand

 $(BAP)=\frac{BA\cdot AP\sin BAP}{2},\qquad(CAP)=\frac{CA\cdot AP\sin CAP}{2}$

and so

 $\frac{(BAP)}{(CAP)}=\frac{BA\cdot AP\sin BAP/2}{CA\cdot AP\sin CAP/2}=\frac{BA% \sin BAP}{CA\sin CAP}.$

We have obtained

 $\frac{BP}{PC}=\frac{BA\sin BAP}{CA\sin CAP},$

which is the generalization to the theorem. In the particular case when $AP$ is the bisector, $\angle BAP=\angle CAP$, and thus $\sin BAP=\sin CAP$. Cancelling out the sines proves the bisector theorem.

Title proof of bisectors theorem ProofOfBisectorsTheorem 2013-03-22 14:49:25 2013-03-22 14:49:25 drini (3) drini (3) 6 drini (3) Proof msc 51A05