If three lines whose equations are $y=m_{1}x+c_{1}$,$y=m_{2}x+c_{2}$,and $y=m_{3}x+c_{3}$ are concurrent,then show that $m_{1}(c_{2}-c_{3})+m_{2}(c_{3}-c_{1})+m_{3}(c_{1}-c_{2})=0$.

(A) The equations of the given lines are:
$y=m_{1}x+c_{1}$ ... $(1)$
$y=m_{2}x+c_{2}$ ... $(2)$
$y=m_{3}x+c_{3}$ ... $(3)$
Subtracting equation $(1)$ from $(2)$,we obtain:
$0 = (m_{2}-m_{1})x + (c_{2}-c_{1})$
$(m_{1}-m_{2})x = c_{2}-c_{1}$
$x = \frac{c_{2}-c_{1}}{m_{1}-m_{2}}$
Substituting this value of $x$ in $(1)$,we get:
$y = m_{1}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right) + c_{1}$
$y = \frac{m_{1}c_{2}-m_{1}c_{1} + m_{1}c_{1}-m_{2}c_{1}}{m_{1}-m_{2}}$
$y = \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}$
Thus,the point of intersection of lines $(1)$ and $(2)$ is $\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}\right)$.
Since the lines are concurrent,this point must satisfy equation $(3)$:
$\frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}} = m_{3}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right) + c_{3}$
$m_{1}c_{2}-m_{2}c_{1} = m_{3}(c_{2}-c_{1}) + c_{3}(m_{1}-m_{2})$
$m_{1}c_{2}-m_{2}c_{1} = m_{3}c_{2}-m_{3}c_{1} + c_{3}m_{1}-c_{3}m_{2}$
$m_{1}(c_{2}-c_{3}) + m_{2}(c_{3}-c_{1}) + m_{3}(c_{1}-c_{2}) = 0$.