Question

a+b/b+c = c+d/d+a

prove that, c=a
a+b+c+d = 0​

Answer 1

Answer:

Hope it will help

Step-by-step explanation:

1st Method:

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

Subtracting 1 from both sides.

(a+b)/(b+c) - 1 = (c+d)/(d+a). - 1

(a+b-b-c)/(b+c) = (c+d-d-a)/(d+a)

(a-c)/(b+c). = (-a+c)/(d+a)

(a-c)/(b+c)+(a-c)/(d+a)=0

(a-c) [ 1/(b+c)+1/(d+a)] = 0

Either. (a-c)=0 => a=c. Proved.

1/(b+c) + 1/(d+a). = 0

1/(b+c). = - 1/(d+a)

d+a= -b-c

a+b+c+d = 0. Proved

2nd Method:

(a+b)/(b+c) = (c+d)/(d+a).

(a+b)/(c+d)= (b+c)/(d+a).

Applying componendo and dividends.

(a+b+c+d)/(a+b-c-d) = (b+c+d+a)/(b+c-d-a).

(a+b+c+d)/(a+b-c-d) = - (a+b+c+d)/(a-b-c+d).

(a+b+c+d).[ 1/(a+b-c-d) + 1/(a-b-c+d)] = 0.

Either (a+b+c+d)=0.

1/(a+b-c-d) + 1/(a-b-c+d) = 0

1/(a+b-c-d) = -1/(a-b-c+d).

-a-b+c+d = a-b-c+d.

c+c = a+a.

2c = 2a

c = a. Proved.

Answer 2

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