solutions of a triangle

**grgrsanjay** · August 11th 2010, 05:22 AM

prove that: $\:\dfrac{c}{a+b}\:\ = \:\dfrac{1-tan\:\dfrac{A}{2}\:\tan\:\dfrac{B}{2}\ }{1 +tan\:\dfrac{A}{2}\:\tan\:\dfrac{B}{2} }$

where $\:\boxed{a,b,c}$ are the sides opposite to the angles $\:\boxed{A,B,C}$ respectively in a triangle

**simplependulum** · August 11th 2010, 05:53 AM

Let $P , Q , R$ be the points of tangency of the incircle to the sides $BC , CA , AB$ respectively .

Then $CP = CQ = (a+b-c)/2 = r \cot(\frac{C}{2})$ where $r$ is the inradius .

Also , $c = AR + BR = r [ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) ]$

But $\cot(\frac{C}{2}) = \tan( \frac{A+B}{2} ) = \tan(\frac{A}{2}) \tan(\frac{B}{2}) ~ \frac{ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) }$

Therefore , $\frac{ a+b-c}{c} = 2r \tan(\frac{A}{2}) \tan(\frac{B}{2}) ~ \frac{ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) } ~ \frac{1}{ r[ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) ]}$

$= \frac{2\tan(\frac{A}{2}) \tan(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) }$

$\frac{ a+b}{c} = \frac{1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) }{1-\tan(\frac{A}{2}) \tan(\frac{B}{2}) }$

or $\boxed{ \displaystyle{ \frac{c}{a+b} = \frac{ 1 -\tan(\frac{A}{2}) \tan(\frac{B}{2}) }{1+ \tan(\frac{A}{2}) \tan(\frac{B}{2}) } }}$