Given equations for the side-lines of a parallelogram, why are these the equations for the diagonal-lines? My book, for a parallelogram $ABCD$ with sides as 
$$\begin{align}
AB&\;\equiv\; a\phantom{^\prime}x+b\phantom{^\prime}y  +c\phantom{^\prime}=0 \\
BC&\;\equiv\; a^\prime x         +b^\prime y           +c^\prime=0 \\ 
CD&\;\equiv\; a\phantom{^\prime}x+b\phantom{^\prime} y +c^\prime=0 \\
DA&\;\equiv\; a^\prime x         +b^\prime y           +c\phantom{^\prime}=0
\end{align}$$
wrote equation of diagonals: 
$$AC\;\equiv\; (ax+by+c)(a'x+b'y+c)-(a'x+b'y+c')(ax+by+c')=0$$
$$BD\;\equiv\; (ax+by+c)(a'x+b'y+c')-(a'x+b'y+c)(ax+by+c')=0$$
I don't understand why. Please help. 
 A: It should be:
$$BD\equiv (ax+by+c)(a'x+b'y+c)-(a'x+b'y+c')(ax+by+c')=0$$$$AC\equiv (ax+by+c)(a'x+b'y+c')-(a'x+b'y+c)(ax+by+c')=0$$
Because  


*

*they are equations of straight lines, 

*$(x_1,y_1)$ is placed on the line $ax+by+c=0$ iff $ax_1+by_1+c=0$ and 

*there is an unique straight line which goes through two distinct points.
Done!
A: These equations are wrong. For example, at the point $A$, the two factors of the first term in the expression for $AC$ are zero, while the remaining term, and so the whole expression, evaluates to $-(c'-c)^2$, which generally is not zero—and similarly for $B$ on $BD$.
A: As Michael Rozenberg has written, they should be
$$\small BD\equiv (ax+by+c)(a'x+b'y+c)-(a'x+b'y+c')(ax+by+c')=0\tag1$$$$\small AC\equiv (ax+by+c)(a'x+b'y+c')-(a'x+b'y+c)(ax+by+c')=0\tag2$$

why the equations of diagonals (taking AC for instance) is for the entire line AC and not just points A and C?

$(1)$ can be written as
$$aa'x^2+ab'xy+acx+a'bxy+bb'y^2+bcy+ca'x+cb'y+c^2-aa'x^2-a'bxy-a'c'x-ab'xy-bb'y^2-b'c'y-ac'x-bc'y-c'^2=0,$$
i.e.
$$(a+a')(c-c')x+(b+b')(c-c')y+(c-c')(c+c')=0\tag3$$
Suppose that $c=c'$. Then, the equations of $AB$ and $CD$ are the same, which is impossible. So, we have $c\not=c'$. 
So, dividing the both sides of $(3)$ by $c-c'$ gives
$$(a+a')x+(b+b')y+c+c'=0\tag4$$
Suppose that $a+a'=0$ and $b+b'=0$. Then, the equation of $DA$ is $ax+by-c=0$. So, the line $DA$ is parallel to the line $AB$, which is impossible. So, we have $a+a'\not=0$ or $b+b'\not=0$.
It follows that $(4)$, i.e. $(1)$ is the equation of the line $BD$ (assuming that you've already known that $B,D$ satisfy the equation).
Also, $(2)$ can be written as$$(a-a')x+(b-b')y=0\tag5$$
Suppose that $a-a'=0$ and $b-b'=0$. Then, the equations of $AB$ and $DA$ are the same, which is impossible. So, we have $a-a'\not=0$ or $b-b'\not=0$.
It follows that $(5)$, i.e. $(2)$ is the equation of the line $AC$.
