Given triangle $ABC$, let $D$ be an inner point of the side $BC$. Let $P$ and $Q$ be distinct inner points of the segment $AD$. Let $K=BP\cap AC$, $L=CP\cap AB$, $E=BQ\cap AC$, $F=CQ\cap AB$.
Given that $KL\parallel EF$, find all possible values of the ratio $BD:DC$.
1. **Claim**: The only possible value of the ratio \( \frac{BD}{DC} \) is \( 1 \).

2. **Verification for \( \frac{BD}{DC} = 1 \)**:
   - When \( BD = DC \), the point \( D \) is the midpoint of \( BC \).
   - By Ceva's Theorem, if \( BD = DC \), then for any points \( P \) and \( Q \) on \( AD \), the lines \( BP \) and \( CP \) will intersect \( AC \) and \( AB \) respectively at points \( K \) and \( L \) such that \( KL \parallel BC \).
   - Similarly, the lines \( BQ \) and \( CQ \) will intersect \( AC \) and \( AB \) respectively at points \( E \) and \( F \) such that \( EF \parallel BC \).
   - Since \( KL \parallel BC \) and \( EF \parallel BC \), it follows that \( KL \parallel EF \).

3. **Contradiction for \( \frac{BD}{DC} \neq 1 \)**:
   - Suppose \( \frac{BD}{DC} \neq 1 \).
   - Let \( X \) be the point on line \( BC \) such that the cross-ratio \((B, C; D, X) = -1\). This means \( X \) is the harmonic conjugate of \( D \) with respect to \( B \) and \( C \).
   - By the properties of harmonic division and Ceva/Menelaus Duality, \( X \) is the point where \( KL \) and \( EF \) intersect \( BC \).
   - Since \( KL \) and \( EF \) intersect at \( X \), they cannot be parallel. This contradicts the given condition that \( KL \parallel EF \).

4. **Conclusion**:
   - The only possible value of the ratio \( \frac{BD}{DC} \) that satisfies the condition \( KL \parallel EF \) is \( 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{BD}{DC} = 1 } \)