# Incircle and Incenter – Length of Angle Bisector AP

## Incircle and Incenter – Length of Angle Bisector AP

Length of Angle Bisector AP:

Area of Triangle ABP + Area of Triangle ACP = Area of Triangle ABC

$$\frac{1}{2}AB\times AP\sin \left( \frac{A}{2} \right)+\frac{1}{2}AC\times AP\ \sin \left( \frac{A}{2} \right)=\frac{1}{2}AB\times AC\sin A$$,

$$\frac{1}{2}\left( c+b \right)AP\times \sin \left( \frac{A}{2} \right)=\frac{1}{2}\left( 2bc\sin \left( \frac{A}{2} \right)\cos \left( \frac{A}{2} \right) \right)$$,

$$AP=\left( \frac{2bc}{b+c} \right)\cos \left( \frac{A}{2} \right)$$

Similarly, length of angle bisector through point B and C is

$$BQ=\left( \frac{2ac}{a+c} \right)\cos \left( \frac{B}{2} \right)$$,

$$CR=\left( \frac{2ab}{a+b} \right)\cos \left( \frac{C}{2} \right)$$,

Example: In Triangle ABC, the three bisectors of the angles A, B and C are extended to incircle at D, E and F respectively. Prove that $$AD\cos \frac{A}{2}+BE\cos \frac{B}{2}+CF\cos \frac{C}{2}=2\left( \sin A+\sin B+\sin C \right)$$.

Solution:

Using sine law in triangle ABD

$$\frac{AD}{\sin \left( B+\frac{A}{2} \right)}=2R$$,

$$AD=2R\sin \left( B+\frac{A}{2} \right)$$,

$$AD\times \cos \frac{A}{2}=2R\cos \frac{A}{2}\sin \left( B+\frac{A}{2} \right)$$,

$$=\sin \left( A+B \right)+\sin B$$,

$$=\sin C+\sin B$$….(1)

Similarly, $$BE\cos \left( \frac{B}{2} \right)=\sin A+\sin C$$,

\begin{align}& AD\cos \left( \frac{A}{2} \right)=\sin A+\sin B \\ & \\ \end{align},

$$AD\cos \left( \frac{A}{2} \right)+BE\cos \left( \frac{B}{2} \right)+CF\cos \left( \frac{C}{2} \right)=2\left( \sin A+\sin B+\sin C \right)$$,

Hence Proved.