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 $A P$ is the bisector, $\\angle BAP=\\angle CAP$ , and thus $\\sin BAP=\\sin CAP$ . Cancelling out the sines proves the bisector theorem.

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