Find all triangles $ABC$ such that $\frac{a cos A + b cos B + c cos C}{a sin A + b sin B + c sin C} =\frac{a + b + c}{9R}$, where, as usual, $a, b, c$ are the lengths of sides $BC, CA, AB$ and $R$ is the circumradius.

1. **Given Equation:**
   \[
   \frac{a \cos A + b \cos B + c \cos C}{a \sin A + b \sin B + c \sin C} = \frac{a + b + c}{9R}
   \]
   where \(a, b, c\) are the lengths of sides \(BC, CA, AB\) respectively, and \(R\) is the circumradius of triangle \(ABC\).

2. **Expression for \(a \cos A + b \cos B + c \cos C\):**
   We use the known identity:
   \[
   a \cos A + b \cos B + c \cos C = \frac{8S^2}{abc}
   \]
   where \(S\) is the area of the triangle.

3. **Expression for \(a \sin A + b \sin B + c \sin C\):**
   Using the Law of Sines, we have:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   Therefore,
   \[
   a \sin A + b \sin B + c \sin C = 2S
   \]
   where \(S\) is the area of the triangle.

4. **Substitute these expressions into the given equation:**
   \[
   \frac{\frac{8S^2}{abc}}{2S} = \frac{a + b + c}{9R}
   \]
   Simplifying, we get:
   \[
   \frac{4S}{abc} = \frac{a + b + c}{9R}
   \]

5. **Relate the area \(S\) to the circumradius \(R\):**
   The area \(S\) of the triangle can also be expressed as:
   \[
   S = \frac{abc}{4R}
   \]
   Substituting \(S\) into the equation, we get:
   \[
   \frac{4 \cdot \frac{abc}{4R}}{abc} = \frac{a + b + c}{9R}
   \]
   Simplifying, we get:
   \[
   \frac{1}{R} = \frac{a + b + c}{9R}
   \]
   Multiplying both sides by \(9R\), we get:
   \[
   9 = a + b + c
   \]

6. **Using the AM-GM inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   a + b + c \geq 3\sqrt[3]{abc}
   \]
   Given \(a + b + c = 9\), we get:
   \[
   9 \geq 3\sqrt[3]{abc}
   \]
   Dividing both sides by 3, we get:
   \[
   3 \geq \sqrt[3]{abc}
   \]
   Cubing both sides, we get:
   \[
   27 \geq abc
   \]

7. **Equality condition in AM-GM:**
   The equality in AM-GM holds if and only if \(a = b = c\). Therefore, for the equality \(a + b + c = 9\) to hold, we must have:
   \[
   a = b = c
   \]

8. **Conclusion:**
   Since \(a = b = c\), the triangle \(ABC\) must be equilateral.

The final answer is \( \boxed{ a = b = c } \).