Given equations for the side-lines of a parallelogram, why are these the equations for the diagonal-lines?

Question

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.

Answer 1

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

1. they are equations of straight lines,

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

3. there is an unique straight line which goes through two distinct points.

Done!

Answer 2

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 $BD$.