Three points P, Q, R lie on the side BC of triangle ABC such that BP = PQ = QR = RC. If G be centroid of `DeltaABC` then what is ratio of areas of `DeltaPGR and DeltaABC`.

Answer

To solve the problem, we need to find the ratio of the areas of triangle PGR to triangle ABC, given that the points P, Q, and R divide the side BC of triangle ABC into equal segments.

### Step-by-Step Solution:

1. **Identify the Points on BC**:
Let the length of segment BC be divided into four equal parts by points P, Q, and R. Therefore, we can denote:
- BP = PQ = QR = RC = x (where x is the length of each segment).
- Thus, the total length BC = 4x.

2. **Determine the Centroid G**:
The centroid G of triangle ABC divides each median in the ratio 2:1. Since G is the centroid, it will also divide the area of triangle ABC into three equal parts.

3. **Area of Triangle ABC**:
Let the area of triangle ABC be denoted as Area(ABC). The centroid divides triangle ABC into three smaller triangles (AGB, AGC, and AGB), each having equal area:
\[ \text{Area}(AGB) = \text{Area}(AGC) = \text{Area}(AGP) = \frac{1}{3} \text{Area}(ABC) \]

4. **Area of Triangle PGR**:
Since P, Q, and R divide BC into equal segments, triangles APG, AQR, and APR will have equal areas. The area of triangle PGR can be calculated as follows:
- The area of triangle APQ is equal to the area of triangle AQR, and since G is the centroid, the area of triangle PGR will be half of the area of triangle AQR.
- Therefore, we can denote the area of triangle PGR as:
\[ \text{Area}(PGR) = \text{Area}(AGP) + \text{Area}(AGR) = \frac{1}{3} \text{Area}(ABC) + \frac{1}{3} \text{Area}(ABC) = \frac{2}{3} \text{Area}(ABC) \]

5. **Calculate the Ratio**:
Now, we can find the ratio of the areas of triangle PGR to triangle ABC:
\[ \text{Ratio} = \frac{\text{Area}(PGR)}{\text{Area}(ABC)} = \frac{2}{3 \cdot 4} = \frac{2}{12} = \frac{1}{6} \]

### Final Answer:
The ratio of the areas of triangle PGR to triangle ABC is \( \frac{1}{6} \).