Let $\Delta$ denote the set of all triangles in a plane. Consider the function $f: \Delta\to(0,\infty)$ defined by $f(ABC) = \min \left( \dfrac ba, \dfrac cb \right)$, for any triangle $ABC$ with $BC=a\leq CA=b\leq AB = c$. Find the set of values of $f$.
1. **Define the function and variables:**
   Let \( \Delta \) denote the set of all triangles in a plane. Consider the function \( f: \Delta \to (0, \infty) \) defined by \( f(ABC) = \min \left( \frac{b}{a}, \frac{c}{b} \right) \), for any triangle \( ABC \) with \( BC = a \leq CA = b \leq AB = c \).

2. **Express the ratios:**
   Define \( \frac{b}{a} = m \) and \( \frac{c}{b} = n \). Given \( a \leq b \leq c \), it follows that \( m, n \geq 1 \).

3. **Triangle inequality constraints:**
   The triangle inequality \( a + b > c \) translates to:
   \[
   a + b > c \implies a + b > \frac{c}{b} \cdot b = n \cdot b \implies 1 + \frac{b}{a} > n \implies 1 + m > mn
   \]

4. **Case analysis:**
   - **Case i: \( m \leq n \)**
     \[
     m \leq n < 1 + \frac{1}{m}
     \]
     Therefore, \( \min\{m, n\} = m \).

   - **Case ii: \( n \leq m \)**
     \[
     n \leq m < \frac{1}{n-1}
     \]
     Therefore, \( \min\{m, n\} = n \).

5. **Combining the cases:**
   In both cases, we have:
   \[
   \min\{m, n\} < \frac{\sqrt{5} + 1}{2}
   \]
   This is because the maximum value of \( \min\{m, n\} \) occurs when \( m = n \), and solving \( 1 + m = m^2 \) gives \( m = \frac{\sqrt{5} + 1}{2} \).

6. **Range of \( f \):**
   We can show that for any \( 0 < \epsilon < \frac{\sqrt{5} - 1}{2} \), there exist positive reals \( m \) and \( n \) such that \( \min\{m, n\} = \frac{\sqrt{5} + 1}{2} - \epsilon \).

7. **Conclusion:**
   Therefore, the set of values of \( f \) is:
   \[
   \left[1, \frac{\sqrt{5} + 1}{2}\right)
   \]

The final answer is \( \boxed{ \left[1, \frac{\sqrt{5} + 1}{2}\right) } \)