Given the parallelogram $ABCD$. The circle $S_1$ passes through the vertex $C$ and touches the sides $BA$ and $AD$ at points $P_1$ and $Q_1$, respectively. The circle $S_2$ passes through the vertex $B$ and touches the side $DC$ at points $P_2$ and $Q_2$, respectively. Let $d_1$ and $d_2$ be the distances from $C$ and $B$ to the lines $P_1Q_1$ and $P_2Q_2$, respectively.
Find all possible values of the ratio $d_1:d_2$.

[i](I. Voronovich)[/i] 
1. **Define the heights and distances:**
   Let \( h_1 \) and \( h_2 \) denote the perpendicular distances from vertex \( C \) to the sides \( AB \) and \( AD \) of the parallelogram \( ABCD \), respectively. These are the heights of the parallelogram from vertex \( C \).

2. **Express \( d_1 \) in terms of \( h_1 \) and \( h_2 \):**
   We need to show that \( d_1 = \sqrt{h_1 h_2} \). To do this, consider the circle \( S_1 \) which passes through \( C \) and touches \( BA \) at \( P_1 \) and \( AD \) at \( Q_1 \). The distance \( d_1 \) is the perpendicular distance from \( C \) to the line \( P_1Q_1 \).

3. **Use the tangency condition:**
   Since \( S_1 \) is tangent to \( BA \) and \( AD \) at \( P_1 \) and \( Q_1 \) respectively, the angles \( \angle CP_1Q_1 \) and \( \angle CQ_1D \) are equal. This implies that:
   \[
   \sin \angle CP_1Q_1 = \sin \angle CQ_1D
   \]

4. **Relate \( d_1 \) to the heights:**
   The distance \( d_1 \) can be expressed as:
   \[
   d_1 = CP_1 \sin \angle CP_1Q_1
   \]
   Since \( \sin \angle CP_1Q_1 = \frac{h_2}{CQ_1} \), we have:
   \[
   d_1 = CP_1 \cdot \frac{h_2}{CQ_1}
   \]

5. **Analogous expression for \( d_1 \):**
   Similarly, we can express \( d_1 \) in another form:
   \[
   d_1 = CQ_1 \cdot \frac{h_1}{CP_1}
   \]

6. **Multiply the two expressions:**
   Multiplying the two expressions for \( d_1 \):
   \[
   d_1^2 = \left( CP_1 \cdot \frac{h_2}{CQ_1} \right) \left( CQ_1 \cdot \frac{h_1}{CP_1} \right) = h_1 h_2
   \]
   Therefore, we have:
   \[
   d_1 = \sqrt{h_1 h_2}
   \]

7. **Apply the same argument for \( d_2 \):**
   By a similar argument, for the circle \( S_2 \) passing through \( B \) and touching \( DC \) at \( P_2 \) and \( Q_2 \), we can show that:
   \[
   d_2 = \sqrt{h_1 h_2}
   \]

8. **Conclude the ratio:**
   Since both \( d_1 \) and \( d_2 \) are equal to \( \sqrt{h_1 h_2} \), the ratio \( d_1 : d_2 \) is:
   \[
   d_1 : d_2 = 1
   \]

The final answer is \( \boxed{ 1 } \)