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

Options

A. `1/2`
B. `3/2`
C. `7/5`
D. `5/3`

Answer

To solve the equation \(\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = ?\) given that \(a, b, c > 0\), we can follow these steps:

### Step 1: Set the common ratio
Let us denote the common ratio as \(k\). Therefore, we can write:
\[ \frac{a}{b+c} = k, \quad \frac{b}{c+a} = k, \quad \frac{c}{a+b} = k \]

### Step 2: Express \(a\), \(b\), and \(c\) in terms of \(k\)
From the first equation, we can express \(a\):
\[ a = k(b+c) \]
From the second equation:
\[ b = k(c+a) \]
From the third equation:
\[ c = k(a+b) \]

### Step 3: Substitute \(a\), \(b\), and \(c\)
Now, we can substitute \(a\) from the first equation into the second and third equations:
1. Substitute \(a = k(b+c)\) into \(b = k(c+a)\):
\[ b = k\left(c + k(b+c)\right) = k(c + kb + kc) \]
Rearranging gives:
\[ b = kc + k^2b + k^2c \]
\[ b - k^2b = kc + k^2c \]
\[ b(1 - k^2) = c(k + k^2) \]
Thus, we have:
\[ b = \frac{c(k + k^2)}{1 - k^2} \]

2. Now substitute \(b\) into the expression for \(c\):
\[ c = k(a+b) = k\left(k(b+c) + b\right) \]

### Step 4: Solve for \(k\)
To find a relationship between \(a\), \(b\), and \(c\), we can assume \(a = b = c\). Let \(a = b = c = x\):
\[ \frac{x}{x+x} = \frac{x}{2x} = \frac{1}{2} \]
Thus, we find that:
\[ k = \frac{1}{2} \]

### Final Result
Therefore, we conclude:
\[ \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = \frac{1}{2} \]