If \(\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b}\), then : (a) \(\frac{x-y}{b-a}=\frac{y-z}{c-b}=\frac{z-x}{a-c}\) (b) \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{y}}{\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{c}}\) (c) \(\frac{x-y}{c}=\frac{y-z}{b}=\frac{z-x}{a}\) (d) none of the above is true

Given the equation:

\[
\frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b} = k \quad (\text{say})
\]

This implies:

\[
x = k(b+c), \quad y = k(c+a), \quad \text{and} \quad z = k(a+b)
\]

From these equations, we can find the differences:

\[
\begin{aligned}
x – y &= k(b+c) – k(c+a) = k(b-a) \\
y – z &= k(c+a) – k(a+b) = k(c-b) \\
z – x &= k(a+b) – k(b+c) = k(a-c)
\end{aligned}
\]

Now, we check option (a):

\[
\frac{x-y}{b-a} = \frac{y-z}{c-b} = \frac{z-x}{a-c}
\]

Substituting the differences we calculated:

\[
\begin{aligned}
\frac{k(b-a)}{b-a} &= \frac{k(c-b)}{c-b} = \frac{k(a-c)}{a-c}
\end{aligned}
\]

Simplifying, we see that each fraction simplifies to \(k\), since the \(b-a\), \(c-b\), and \(a-c\) in the numerators and denominators cancel out:

\[
k = k = k
\]

Therefore, option (a) \(\frac{x-y}{b-a} = \frac{y-z}{c-b} = \frac{z-x}{a-c}\) is true based on the given equation.