<s>[INST] If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ . 
If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ .

What I Tried :- First I thought for a moment and found out that I can write this :
$$\frac{(a-c)(b-d)}{(b-c)(d-a)} = \frac{(a-c)(b-d)}{(a-b)(c-d)} * \frac{2016}{2017}$$
But how is it going to help?
Then I thought maybe cross-multiply everything and try to factor it? I get :-
$$2017(a-b)(c-d) = 2016(b-c)(d-a)$$
$$\rightarrow 2017(a-b)(c-d) - 2016(b-c)(d-a) = 0$$
$$\rightarrow 2017(ac - ad - bc + bd) - 2016(bd - ab - cd - ac) = 0$$
$$\rightarrow 2017ac + bd + 2016ab + 2016cd + 2016ac - 2017ad - 2017bc - 2016bd = 0$$
I am hopeless at this point, how do you even factorise this?
Can anyone help?
 [/INST] Let
$$t=\frac{(a-b)(c-d)}{(b-c)(d-a)}=\frac{ac+bd-ad-bc}{ac+bd-ab-cd}$$
then
$$t-1=\frac{ab+cd-ad-bc}{ac+bd-ab-cd}=\frac{(a-c)(b-d)}{(b-c)(d-a)}$$
so
$$\frac{(a-c)(b-d)}{(a-b)(c-d)}=\frac{t-1}{t}.$$
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