If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ .

Question

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?

Answer 1

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 $$\frac{(a-c)(b-d)}{(a-b)(c-d)}=\frac{t-1}{t}.$$

Answer 2

Quite similar approach:

$$\frac{2016}{2017}=1-\frac{1}{2017}$$

$$\frac{(a-b)(c-d)}{(b-c)(d-a)}=1-\frac{1}{(b-c)(d-a)}$$

After reducing we get:

$$(a-c)(b-d)=-1=-\frac{(a-b)(c-d)}{(a-b)(c-d)}$$

Or:

$$\frac{(a-c)(b-d)}{(a-b)(c-d)}=\frac{-1}{(a-b)(c-d)}=-\frac{1}{2016}$$