Let $ABC$ be an acute triangle and let $X$ be a variable point on $AC$. The incircle of $\triangle ABX$ touches $AX, BX$ at $K, P$, respectively. The incircle of $\triangle BCX$ touches $CX, BX$ at $L, Q$, respectively. Find the locus of $KP \cap LQ$.
1. **Setup and Definitions:**
   - Let \( \triangle ABC \) be an acute triangle.
   - Let \( X \) be a variable point on \( AC \).
   - The incircle of \( \triangle ABX \) touches \( AX \) and \( BX \) at \( K \) and \( P \), respectively.
   - The incircle of \( \triangle BCX \) touches \( CX \) and \( BX \) at \( L \) and \( Q \), respectively.

2. **Incircle Properties:**
   - The incircle of a triangle touches the sides at points where the angle bisectors intersect the sides.
   - For \( \triangle ABX \), the incircle touches \( AX \) at \( K \) and \( BX \) at \( P \).
   - For \( \triangle BCX \), the incircle touches \( CX \) at \( L \) and \( BX \) at \( Q \).

3. **Intersection Points:**
   - Let \( M \) and \( N \) be the midpoints of \( AB \) and \( AC \), respectively.
   - Let \( I \) be the incenter of \( \triangle ABC \).
   - Let \( BI \cap MN = Z \) and \( CI \cap MN = Y \).

4. **Iran Lemma Application:**
   - The Iran lemma states that for a triangle with an incircle, the intersection of the angle bisectors with the opposite sides lies on the line connecting the midpoints of the other two sides.
   - Applying the Iran lemma, we find that \( KP \cap LQ \) lies on the line \( YZ \).

5. **Angle Chasing:**
   - To confirm the locus, we perform an angle chase:
     - Consider the angles formed by the incircles and the sides of the triangles.
     - The points \( K, P, L, Q \) are determined by the tangency points of the incircles.
     - The intersection \( KP \cap LQ \) must lie on the line \( YZ \) due to the symmetry and properties of the incircles and the angle bisectors.

6. **Conclusion:**
   - The locus of \( KP \cap LQ \) is the line segment \( YZ \).

\(\blacksquare\)

The final answer is the line segment \( \boxed{ YZ } \).