2015/050) If $ax + y + 1= 0, x +by+1=0,$ $x+y+c=0$

are concurrent then prove that $\dfrac{1}{1-a} + \dfrac{1}{1-b} + \dfrac{1}{1-c} = 1$

proof

We have
$a = - \dfrac{y+1}{x}$
or $1- a = \dfrac{x+y+1}{x}$
or $\dfrac{1}{1-a} = \dfrac{x}{x+y + 1} \cdots 1$

similarly

$x +by+1=0$
=> $x+1 = - by$
or $b = -\dfrac{x+1}{y}$
or $1-b= \dfrac{x+y+1}{y}$
or $\dfrac{1}{1-b} = \dfrac{y}{x+y+1}\cdots (2)$

and $x+y+c=0$
=> $-c = x + y$
=> $1-c = x+y+1$
or $\dfrac{1}{1-c} = \dfrac{1}{x+y+1}\cdots(3)$

adding all 3 we get the
$\dfrac{1}{1-a} + \dfrac{1}{1-b} + \dfrac{1}{1-c} = \dfrac{x+ y + 1}{x+y+1} =1$

Proved