An equilateral triangle has been drawn inside the circle. Split the triangle to two parts with equal area by a line segment parallel to the triangle side. Draw an inscribed circle inside this smaller triangle. What is the ratio of the area of this circle compared to the area of original circle.
1. **Determine the side length of the larger equilateral triangle:**
   - Let the radius of the larger circle be \( R \).
   - The side length \( s \) of an equilateral triangle inscribed in a circle of radius \( R \) is given by:
     \[
     s = \sqrt{3}R
     \]

2. **Calculate the area of the larger equilateral triangle:**
   - The area \( A \) of an equilateral triangle with side length \( s \) is:
     \[
     A = \frac{\sqrt{3}}{4} s^2
     \]
   - Substituting \( s = \sqrt{3}R \):
     \[
     A = \frac{\sqrt{3}}{4} (\sqrt{3}R)^2 = \frac{\sqrt{3}}{4} \cdot 3R^2 = \frac{3\sqrt{3}}{4} R^2
     \]

3. **Determine the side length of the smaller equilateral triangle:**
   - The smaller triangle has half the area of the larger triangle:
     \[
     \frac{3\sqrt{3}}{4} R^2 = 2 \cdot \frac{\sqrt{3}}{4} a^2
     \]
   - Solving for \( a \):
     \[
     \frac{3\sqrt{3}}{4} R^2 = \frac{\sqrt{3}}{4} a^2 \implies 3R^2 = a^2 \implies a = \sqrt{3}R
     \]

4. **Calculate the radius of the incircle of the smaller triangle:**
   - The radius \( r \) of the incircle of an equilateral triangle with side length \( a \) is given by:
     \[
     r = \frac{a \sqrt{3}}{6}
     \]
   - Substituting \( a = \sqrt{1.5}R \):
     \[
     r = \frac{\sqrt{1.5}R \cdot \sqrt{3}}{6} = \frac{\sqrt{4.5}R}{6} = \frac{R}{2\sqrt{2}}
     \]

5. **Calculate the area of the incircle of the smaller triangle:**
   - The area of the incircle is:
     \[
     A_{\text{incircle}} = \pi r^2 = \pi \left( \frac{R}{2\sqrt{2}} \right)^2 = \pi \frac{R^2}{8}
     \]

6. **Calculate the area of the original circle:**
   - The area of the original circle is:
     \[
     A_{\text{original}} = \pi R^2
     \]

7. **Determine the ratio of the areas:**
   - The ratio of the area of the incircle of the smaller triangle to the area of the original circle is:
     \[
     \frac{A_{\text{incircle}}}{A_{\text{original}}} = \frac{\pi \frac{R^2}{8}}{\pi R^2} = \frac{1}{8}
     \]

The final answer is \(\boxed{\frac{1}{8}}\).