Let $ABC$ be an acute triangle. Let$H$ and $D$ be points on $[AC]$ and $[BC]$, respectively, such that $BH \perp AC$ and $HD \perp BC$. Let $O_1$ be the circumcenter of $\triangle ABH$, and $O_2$ be the circumcenter of $\triangle BHD$, and $O_3$ be the circumcenter of $\triangle HDC$. Find the ratio of area of $\triangle O_1O_2O_3$ and $\triangle ABH$.
1. **Identify the circumcenters**:
   - \( O_1 \) is the circumcenter of \( \triangle ABH \).
   - \( O_2 \) is the circumcenter of \( \triangle BHD \).
   - \( O_3 \) is the circumcenter of \( \triangle HDC \).

2. **Determine the positions of \( O_1, O_2, \) and \( O_3 \)**:
   - Since \( BH \perp AC \) and \( HD \perp BC \), \( H \) is the orthocenter of \( \triangle ABC \).
   - \( O_1 \) is the midpoint of the hypotenuse \( AB \) in \( \triangle ABH \).
   - \( O_2 \) is the midpoint of the hypotenuse \( BH \) in \( \triangle BHD \).
   - \( O_3 \) is the midpoint of the hypotenuse \( HC \) in \( \triangle HDC \).

3. **Calculate the lengths of segments**:
   - \( O_1O_2 \) is half the length of \( AH \) because \( O_1 \) and \( O_2 \) are midpoints.
   - \( O_2O_3 \) is half the length of \( BC \) because \( O_2 \) and \( O_3 \) are midpoints.

4. **Establish the relationship between the triangles**:
   - Extend \( O_1O_2 \) until it intersects \( AB \) at \( E \). Then \( \triangle EO_1O_2 \) and \( \triangle BAC \) are similar.
   - Denote the heights from \( O_1 \) and \( A \) as \( h_1 \) and \( h_A \), respectively.

5. **Calculate the ratio of heights**:
   - The ratio of the heights is given by:
     \[
     \frac{h_1}{h_A} = \frac{O_1O_2}{AC} = \frac{1}{2} \cdot \frac{AH}{AC}
     \]

6. **Calculate the areas of the triangles**:
   - The area of \( \triangle O_1O_2O_3 \) is:
     \[
     \text{Area}(O_1O_2O_3) = \frac{1}{2} \cdot h_1 \cdot O_2O_3
     \]
   - The area of \( \triangle ABH \) is:
     \[
     \text{Area}(ABH) = \frac{1}{2} \cdot BH \cdot AH
     \]

7. **Substitute the values**:
   - Using the relationships established:
     \[
     \frac{\text{Area}(O_1O_2O_3)}{\text{Area}(ABH)} = \frac{\frac{1}{2} \cdot h_1 \cdot O_2O_3}{\frac{1}{2} \cdot BH \cdot AH} = \frac{h_1 \cdot O_2O_3}{BH \cdot AH}
     \]
   - Substitute \( O_2O_3 = \frac{BC}{2} \) and \( h_1 = \frac{1}{2} \cdot \frac{AH}{AC} \cdot h_A \):
     \[
     \frac{\text{Area}(O_1O_2O_3)}{\text{Area}(ABH)} = \frac{\left(\frac{1}{2} \cdot \frac{AH}{AC} \cdot h_A\right) \cdot \frac{BC}{2}}{BH \cdot AH} = \frac{\frac{1}{4} \cdot AH \cdot BC}{BH \cdot AH} = \frac{1}{4}
     \]

The final answer is \(\boxed{\frac{1}{4}}\).