Question

If
ad - bc/a-b-c+d=ac-bd/a-b-d+c then show that each ratio is equal to a+b+c+d/4.Correct answer would be marked as brainliest and given 30 points but wrong answer will be report and block.So if any one dare solve it
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Answer 1

Answer:

(ad-bc)/(a-b-c+d) = (ac-bd)/(a-b-d+c) = k (let)

(ad-bc). = (a-b-c+d)k.........(1)

(ac-bd). = (a-b-d+c)k .......(2)

adding the two

(ac+ad-bc-bd) = (2a-2b)k

(a(c+d)-b(c+d))= 2k(a-b)

(a-b)(c+d) = 2k(a-b)

(a-b)((c+d) -2k)=0

this gives

either a=b or (c+d) = 2k........(3)

now subtracting (1) from (2) we get

(ac-ad + bc-bd) = (2c-2d)k

(a(c-d)+b(c-d)) = 2k(c-d)

(c-d)(a+b) - 2k(c-d) = 0

(c-d)((a+b) -2k)= 0

this gives

either c= d, or (a+b) = 2k.......(4)

from (3) and (4) we can conclude that

a=b=c=d, however, in that case our expression will become of the form of 0/0 which is not defined. so, a is not equal to b and c is not equal to d.

hence, other solution

c+d = 2k

a+b = 2k

adding

4k = a+b+c+d

k = (a+b+c+d)/4

hence, each of the two ratios is equal to (a+b+c+d)/4