=> \(\dfrac{a+b+c+d}{a-b+c-d}=\dfrac{a+b-c-d}{a-b-c+d}\)
=> \(\dfrac{\left(a+b+c+d\right)+\left(a+b-c-d\right)}{\left(a-b+c-d\right)+\left(a-b-c+d\right)}=\dfrac{\left(a+b+c+d-\left(a+b-c-d\right)\right)}{\left(a-b+c-d\right)-\left(a-b-c+d\right)}\)
=> \(\dfrac{2a+2b}{2a-2b}=\dfrac{2c+2d}{2c-2d}\)
=>\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
=> (a+b)(c-d) = (a-b)(c+d)
=> ac-ad+bc-bd = ac+ac-bc-bd
=>bc-ad = ad-bc
=> bc-ad-ad+bc = 0
=>2(bc-ad) = 0
=> bc- ad = 0
=> bc = ad
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)