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.