`(a)/(b+c)=(b)/(c+a)=(c )/(a+b)=?`
if `a,b,c, gt 0`

Answer

To solve the equation \(\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = k\), where \(k\) is a constant, we can follow these steps:

### Step 1: Set up the equations
From the given ratios, we can express \(a\), \(b\), and \(c\) in terms of \(k\):
\[ a = k(b+c) \]
\[ b = k(c+a) \]
\[ c = k(a+b) \]

### Step 2: Substitute to find relationships
Substituting \(a\) from the first equation into the second:
\[ b = k(c + k(b+c)) \]
This simplifies to:
\[ b = kc + k^2b + k^2c \]
Rearranging gives:
\[ b - k^2b = kc + k^2c \]
Factoring out \(b\):
\[ b(1 - k^2) = c(k + k^2) \]
Thus, we have:
\[ b = \frac{c(k + k^2)}{1 - k^2} \quad \text{(1)} \]

### Step 3: Substitute \(b\) into the equation for \(c\)
Now we substitute \(b\) from equation (1) into the equation for \(c\):
\[ c = k(a + b) \]
Substituting \(b\):
\[ c = k\left(a + \frac{c(k + k^2)}{1 - k^2}\right) \]
This simplifies to:
\[ c = ka + \frac{kc(k + k^2)}{1 - k^2} \]
Rearranging gives:
\[ c(1 - \frac{k(k + k^2)}{1 - k^2}) = ka \]
This can be further simplified to find a relationship between \(a\), \(b\), and \(c\).

### Step 4: Solve for \(k\)
We can also set all three equations equal to \(k\) and solve for \(k\):
\[ \frac{a}{b+c} = \frac{b}{c+a} \implies a(c+a) = b(b+c) \]
Expanding both sides gives:
\[ ac + a^2 = b^2 + bc \]
Rearranging gives:
\[ a^2 - b^2 + ac - bc = 0 \]
This is a quadratic equation in terms of \(a\) and \(b\).

### Step 5: Find the value of \(k\)
By symmetry, we can assume \(a = b = c\). Let \(a = b = c = x\):
\[ \frac{x}{x+x} = \frac{x}{2x} = \frac{1}{2} \]
Thus, we find:
\[ k = \frac{1}{2} \]

### Final Result
Therefore, the value of \(\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = \frac{1}{2}\).

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