Let $ABC$ be an equilateral triangle. $A $ point $P$ is chosen at random within this triangle. What is the probability that the sum of the distances from point $P$ to the sides of triangle $ABC$ are measures of the sides of a triangle?
1. **Understanding the Problem:**
   We are given an equilateral triangle \(ABC\) and a point \(P\) chosen randomly within this triangle. We need to find the probability that the sum of the distances from point \(P\) to the sides of triangle \(ABC\) are measures of the sides of a triangle.

2. **Key Insight:**
   In an equilateral triangle, the sum of the perpendicular distances from any interior point to the three sides is equal to the altitude of the triangle. This is a well-known geometric property.

3. **Altitude Calculation:**
   Let the side length of the equilateral triangle be \(a\). The altitude \(h\) of an equilateral triangle is given by:
   \[
   h = \frac{a\sqrt{3}}{2}
   \]

4. **Condition for Triangle Inequality:**
   For the distances \(d_1, d_2, d_3\) from point \(P\) to the sides of the triangle to form the sides of a triangle, they must satisfy the triangle inequality:
   \[
   d_1 + d_2 > d_3, \quad d_2 + d_3 > d_1, \quad d_3 + d_1 > d_2
   \]
   Given that \(d_1 + d_2 + d_3 = \frac{a\sqrt{3}}{2}\), we need to ensure that none of these distances is more than half of the altitude, i.e., \(\frac{a\sqrt{3}}{4}\).

5. **Geometric Interpretation:**
   The condition \(d_1, d_2, d_3 \leq \frac{a\sqrt{3}}{4}\) implies that point \(P\) must lie within a smaller equilateral triangle formed by connecting the midpoints of the sides of the original triangle \(ABC\).

6. **Area Calculation:**
   The smaller equilateral triangle formed by the midpoints of the sides of \(ABC\) has side length \(\frac{a}{2}\). The area of this smaller triangle is:
   \[
   \text{Area of smaller triangle} = \frac{\sqrt{3}}{4} \left(\frac{a}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{\sqrt{3}a^2}{16}
   \]
   The area of the original triangle \(ABC\) is:
   \[
   \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} a^2
   \]

7. **Probability Calculation:**
   The probability that point \(P\) lies within the smaller triangle is the ratio of the area of the smaller triangle to the area of the original triangle:
   \[
   \text{Probability} = \frac{\text{Area of smaller triangle}}{\text{Area of } \triangle ABC} = \frac{\frac{\sqrt{3}a^2}{16}}{\frac{\sqrt{3}a^2}{4}} = \frac{1}{4}
   \]

Conclusion:
\[
\boxed{\frac{1}{4}}
\]