proof of bisectors theorem

Notice that the triangles $\mathrm{△}BAP$ and $\mathrm{△}CAP$ 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

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

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, $\mathrm{\angle }BAP=\mathrm{\angle }CAP$, and thus $\sin BAP=\sin CAP$. Cancelling out the sines proves the bisector theorem.

Title	proof of bisectors theorem
Canonical name	ProofOfBisectorsTheorem
Date of creation	2013-03-22 14:49:25
Last modified on	2013-03-22 14:49:25
Owner	drini (3)
Last modified by	drini (3)
Numerical id	6
Author	drini (3)
Entry type	Proof
Classification	msc 51A05