2016/041) If $ax + y + 1= 0, x +by+1=0,$ $x+y+c=0$ are concurrent then prove that $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

proof
We have
$a = - \frac{y+1}{x}$
or $1- a = \frac{x+y+1}{x}$
or $\frac{1}{1-a} = \frac{x}{x+y + 1} \cdots 1$
Similarly
or $x +by+1=0$
or  $x+1 = - by$
or $b = -\frac{x+1}{y}$
or $1-b= \frac{x+y+1}{y}$
or $\frac{1}{1-b} = \frac{y}{x+y+1}\cdots (2)$

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

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

Proved