Switching some Variables

Let $(a-b)=p\,\,; (c-d)=q\,\,; (b-c)=r\,\, ; (d-a)=s$

$a-b+b-c=a-c=p+r$ and $c-d+d-a=c-a=q+s$

So, $a-c+c-a=p+q+r+s=0$

Now put these values in the given fraction $\dfrac{pq}{rs}=-\dfrac{4}{7}$

The fraction which is need to be evaluated becomes $\dfrac{(p+r)(q+r)}{pq} \\ = \dfrac{pq+r(p+q) + r^2}{pq} \\= \dfrac{pq+r(-r-s) +r^2}{pq}...........As\,\, (p+q+r+s)=0 \\ =\dfrac{pq -rs}{pq} \\ =1-\dfrac{rs}{pq} =\dfrac{11}{4}$

Given that $\dfrac{(a-b)(c-d)}{(b-c)(d-a)} = -\dfrac{4}{7}$ .

Let's do some manipulations.

$\begin{aligned} \dfrac{(a-b)(c-d)}{(b-c)(d-a)} & = -\dfrac{4}{7} \\ 7(a-b)(c-d) & = -4(b-c)(d-a) \\ 7(ac - ad - bc + bd ) & = -4(bd - ab - cd + ac) \\ 7ac - 7ad - 7bc + 7 bd & = -4bd + 4ab + 4cd - 4ac \quad \color{#3D99F6} \Rightarrow \text{Adding both sides with } \space 4ac - 4ad -4bc + 4bd \\ \color{#333333} 11ac - 11ad - 11bc + 11bd & = 4ab - 4ad -4bc + 4cd \\ 11(ac - ad - bc + bd) & = 4( ab - ad - bc + cd) \\ 11(a-b)(c-d) & = 4(a-c)(b-d) \\ \dfrac{(a-c)(b-d)}{(a-b)(c-d)} & = \dfrac{11}{4} = \dfrac{A}{B} \end{aligned}$

We got $A= 11, \space B = 4$ . Then, $\dfrac{A+B}{A-B} = \dfrac{15}{7} = \dfrac{C}{D}$ .

We got $C = 15, \space D = 7$ .

Hence, $A+B+C+D = 37$ .

$Applying~componendo-dividendo,:-\\ \dfrac{(a-b)(c-d)}{(b-c)(d-a)} = -\dfrac{4}{7},~~\implies~~\dfrac{ac+bd-bc-ad}{bd+ac-ab-cd}=-\dfrac {-4} 7,\\ \therefore~\dfrac{-ad-bc+ab+cd}{bd+ac-ab-cd}=\dfrac{-4-7} 7.......Using~~\dfrac p q=\dfrac r s, ~~\therefore~\dfrac {p-q} q=\dfrac{r-s} s.\\ and ~~ \dfrac{-ad-bc+ab+cd}{bd+ac-ad-bc}=\dfrac{-11} {-4}..........Using~~\dfrac p q=\dfrac r s, ~~\therefore~\dfrac {p+q} q=\dfrac{r+s} s.\\ Factorizing~~\dfrac{(a-c)(b-d)}{(a-d)(c-b)}=\dfrac{11} 4.\\ So~~\dfrac A B=\dfrac{11} 4~~\\ and~~so~~\dfrac{A+B}{A-B}=\dfrac{11+4}{11-4}=\dfrac {15} 7=\dfrac C D.........Using~~ \dfrac p q=\dfrac r s,~~\therefore~\dfrac{p+q}{p-q}=\dfrac{r+s}{r-s}..\\ \implies~~A+B+C+D=11+4+15+7=\Large~~\color{#D61F06}{37}.$

$\alpha =\quad \frac { (a-b)(c-d) }{ (b-c)(d-a) } \Leftrightarrow \quad \frac { 1 }{ \alpha } =\quad \frac { (b-c)(d-a) }{ (a-b)(c-d) } \\ \beta =\quad \frac { (a-c)(b-d) }{ (a-b)(c-d) } \\ \therefore \quad \frac { 1 }{ \alpha } +\beta \quad =\frac { bd-ab-cd+ac+ab-ad-bc+cd }{ (a-b)(c-d) } =\frac { (a-b)(c-d) }{ (a-b)(c-d) } \quad =1\\ \Rightarrow \quad \beta =\quad 1-\frac { 1 }{ \alpha } =\quad 1+\frac { 7 }{ 4 } =\boxed { \frac { 11 }{ 4 } }$