Point $O$ is chosen in a triangle $ABC$ such that ${d_a},{d_b},{d_c}$ are distance from point $O$ to sides $BC,CA,AB$, respectively. Find position of point $O$ so that product ${d_a} \cdot {d_b} \cdot {d_c}$ becomes maximum.
To find the position of point \( O \) in triangle \( ABC \) such that the product \( d_a \cdot d_b \cdot d_c \) is maximized, we can use the following steps:

1. **Express the distances in terms of the area of the triangle:**
   Let \( \Delta \) be the area of triangle \( ABC \). The distances from point \( O \) to the sides \( BC \), \( CA \), and \( AB \) are \( d_a \), \( d_b \), and \( d_c \) respectively. The area of triangle \( ABC \) can be expressed as:
   \[
   \Delta = \frac{1}{2} \cdot a \cdot d_a = \frac{1}{2} \cdot b \cdot d_b = \frac{1}{2} \cdot c \cdot d_c
   \]
   Therefore, we have:
   \[
   d_a = \frac{2\Delta}{a}, \quad d_b = \frac{2\Delta}{b}, \quad d_c = \frac{2\Delta}{c}
   \]

2. **Formulate the product \( d_a \cdot d_b \cdot d_c \):**
   \[
   d_a \cdot d_b \cdot d_c = \left( \frac{2\Delta}{a} \right) \left( \frac{2\Delta}{b} \right) \left( \frac{2\Delta}{c} \right) = \frac{8\Delta^3}{abc}
   \]

3. **Apply the AM-GM inequality:**
   To maximize \( d_a \cdot d_b \cdot d_c \), we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For any non-negative real numbers \( x, y, z \):
   \[
   \frac{x + y + z}{3} \geq \sqrt[3]{xyz}
   \]
   Equality holds if and only if \( x = y = z \).

4. **Apply the inequality to our problem:**
   Let \( x = ad_a \), \( y = bd_b \), and \( z = cd_c \). Then:
   \[
   \frac{ad_a + bd_b + cd_c}{3} \geq \sqrt[3]{(ad_a)(bd_b)(cd_c)}
   \]
   Since \( ad_a + bd_b + cd_c = 2\Delta \), we have:
   \[
   \frac{2\Delta}{3} \geq \sqrt[3]{(ad_a)(bd_b)(cd_c)}
   \]
   Cubing both sides:
   \[
   \left( \frac{2\Delta}{3} \right)^3 \geq (ad_a)(bd_b)(cd_c)
   \]
   \[
   \frac{8\Delta^3}{27} \geq (ad_a)(bd_b)(cd_c)
   \]

5. **Determine the condition for equality:**
   Equality holds if and only if \( ad_a = bd_b = cd_c \). This condition is satisfied when point \( O \) is the centroid of triangle \( ABC \). The centroid divides each median in the ratio 2:1, and it is the point where the medians intersect.

Conclusion:
The product \( d_a \cdot d_b \cdot d_c \) is maximized when point \( O \) is the centroid of triangle \( ABC \).

The final answer is the centroid of the triangle \( \boxed{ ABC } \).