Ta có hệ phương trình:
$\begin{cases} \frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} \\ a+b+c \end{cases}$

Gọi $k$ là giá trị của các tỉ số bằng nhau, ta có:
$\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{a+b} = k$

Từ đó, ta có:
$\begin{cases} a = k(b+c) \\ b = k(c+a) \\ c = k(a+b) \end{cases}$

Thay các giá trị của $a, b, c$ vào phương trình $a+b+c$, ta có:
$k(b+c) + k(c+a) + k(a+b) = a+b+c$

Simplifying the equation, we get:
$2k(a+b+c) = a+b+c$

Since $a+b+c$ is non-zero, we can divide both sides of the equation by $a+b+c$:
$2k = 1$

Thus, $k = \frac{1}{2}$.

Substituting this value of $k$ back into the equations, we get:
$\begin{cases} a = \frac{1}{2}(b+c) \\ b = \frac{1}{2}(c+a) \\ c = \frac{1}{2}(a+b) \end{cases}$

Simplifying these equations, we get:
$\begin{cases} a = \frac{1}{2}b + \frac{1}{2}c \\ b = \frac{1}{2}c + \frac{1}{2}a \\ c = \frac{1}{2}a + \frac{1}{2}b \end{cases}$

Adding these equations together, we get:
$a + b + c = \frac{3}{2}(a + b + c)$

Dividing both sides by $\frac{3}{2}$, we get:
$a + b + c = 2(a + b + c)$

Since $a+b+c$ is non-zero, we can divide both sides of the equation by $a+b+c$:
$2 = 1$

This is a contradiction, so there are no solutions for $a, b, c$ that satisfy the given conditions.