Question

(b+c) (b+d)=(c+a)(c+d) proved in 'k' formula ?​

Answer 1

Answer:

answer in attachment

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given a,b,c,d are in continued proportion

⟹

b

a

=

c

b

=

d

c

=k(say)

⟹c=dk,b=ck=k

2

d,a=bk=k

3

d

LHS=(a+d)(b+c)−(a+c)(b+d)=(k

3

d+d)(k

2

d+kd)−(k

3

d+kd)(k

2

d+d)

=d

2

(k

5

+k

2

+k

4

+k−k

5

−2k

3

−k)

=k

2

d

2

(k−1)

2

=(k

2

d−kd)

2

=(b−c)

2

=RHS

Hence Proved