a-b-c-and-b-c-a-Prove-that-c-a-b-

Question Number 215973 by MATHEMATICSAM last updated on 23/Jan/25

(a + b) \propto c and (b + c) \propto a . Prove that

(c + a) \propto b .

Answered by Rasheed.Sindhi last updated on 24/Jan/25

(a + b) \propto c and (b + c) \propto a . Prove that

(c + a) \propto b .

\frac{a + b}{c} = k_{1}, \frac{b + c}{a} = k_{2}

\frac{a + b}{c} + 1 = k_{1} + 1, \frac{b + c}{a} + 1 = k_{2} + 1

\frac{a + b + c}{c} = k_{1} + 1, \frac{a + b + c}{a} = k_{2} + 1

a + b + c = {c k}_{1} + c = {a k}_{2} + a

c = \frac{{a k}_{2} + a}{k_{1} + 1}

c + a = \frac{{a k}_{2} + a}{k_{1} + 1} + a = \frac{{a k}_{2} + a + {a k}_{1} + a}{k_{1} + 1}

c + a = a (\frac{k_{1} + k_{2} + 2}{k_{1} + 1})

\Rightarrow c + a \propto a ? ?

\dots .

Commented by MATHEMATICSAM last updated on 24/Jan/25

Can you try by cancelling a + b + c by

dividing and making a = mc (m = const)

Commented by Rasheed.Sindhi last updated on 24/Jan/25

a = mc \Rightarrow a \propto c

why should we assume a \propto c ?

Answered by Rasheed.Sindhi last updated on 24/Jan/25

a + b = p c \dots . . (i)

b + c = q a \dots . (i i)

l e t c + a = r b; r i s c o n s t a n t \dots . . (i i i)

(i i) - (i) :

c - a = q a - p c \dots . (i v)

(i i i) + (i v) :

2 c = r b + q a - p c \dots (v)

(i i i) - (i v) :

2 a = r b - q a + p c \dots . (v i)

(v) + (v i) :

2 c + 2 a = 2 r b \Rightarrow c + a = r b \Rightarrow c + a \propto b

Commented by Rasheed.Sindhi last updated on 24/Jan/25

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