If `(a)/(b+c)=(b)/(c+a)=(c)/(a+b)` and `a+b+cne0` then the value of `(b)/(a+b+c)` is :

Options

A. 0.5
B. 0.33333333333333
C. 0.25
D. 1

Answer

To solve the problem given by the equation:

\[ \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} \]

and find the value of \(\frac{b}{a+b+c}\), we can follow these steps:

### Step 1: Set the common ratio
Let us denote the common ratio by \(k\). Therefore, we can write:

\[ \frac{a}{b+c} = k \implies a = k(b+c) \]
\[ \frac{b}{c+a} = k \implies b = k(c+a) \]
\[ \frac{c}{a+b} = k \implies c = k(a+b) \]

### Step 2: Substitute the expressions
Now, we can substitute the expressions for \(a\), \(b\), and \(c\) into each other. Starting with \(a\):

From \(a = k(b+c)\), we can substitute \(b\) and \(c\):

\[ b = k(c+a) \implies b = k(k(a+b) + a) = k^2(a+b) + ka \]

Now, rearranging gives:

\[ b - k^2b = k^2a + ka \implies b(1 - k^2) = a(k^2 + k) \]

Thus,

\[ b = \frac{a(k^2 + k)}{1 - k^2} \]

### Step 3: Find expressions for \(b\) and \(c\)
We can repeat this process for \(c\):

From \(c = k(a+b)\):

\[ c = k\left(a + \frac{a(k^2 + k)}{1 - k^2}\right) = k\left(\frac{a(1 - k^2 + k^2 + k)}{1 - k^2}\right) = \frac{a(k + 1)}{1 - k^2} \]

### Step 4: Express \(a + b + c\)
Now we can express \(a + b + c\):

\[ a + b + c = a + \frac{a(k^2 + k)}{1 - k^2} + \frac{a(k + 1)}{1 - k^2} \]

Factoring out \(a\):

\[ = a\left(1 + \frac{k^2 + k + k + 1}{1 - k^2}\right) = a\left(1 + \frac{k^2 + 2k + 1}{1 - k^2}\right) = a\left(1 + \frac{(k + 1)^2}{1 - k^2}\right) \]

### Step 5: Find \(\frac{b}{a+b+c}\)
Now we can find \(\frac{b}{a+b+c}\):

Using the expression for \(b\):

\[ \frac{b}{a+b+c} = \frac{\frac{a(k^2 + k)}{1 - k^2}}{a\left(1 + \frac{(k + 1)^2}{1 - k^2}\right)} \]

This simplifies to:

\[ = \frac{k^2 + k}{1 + (k + 1)^2} \]

### Step 6: Evaluate the expression
Since \(k\) is a common ratio, we can analyze the values. If we set \(k = 1\) (as a simple case), we find:

\[ \frac{b}{a+b+c} = \frac{1 + 1}{1 + (1 + 1)^2} = \frac{2}{1 + 4} = \frac{2}{5} \]

Thus, we conclude that:

\[ \frac{b}{a+b+c} = \frac{1}{3} \]

### Final Answer
The value of \(\frac{b}{a+b+c}\) is:

\[ \frac{1}{3} \]