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To solve the problem, let's carefully analyze the given information and proceed step by step.

Problem Statement:

We are given a triangle \( \triangle ABC \) with an inscribed circle (incircle) that touches the sides \( BC \), \( CA \), and \( AB \) at points \( D \), \( E \), and \( F \), respectively. The incircle has center \( I \). A line passing through \( I \) intersects the segments \( AD \) and \( EF \) at points \( K \) and \( L \), respectively. We are tasked with proving that \( IK = IL \).

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Step-by-Step Solution:

#### 1. Understanding the Geometry:
- The incircle of \( \triangle ABC \) is tangent to \( BC \), \( CA \), and \( AB \) at points \( D \), \( E \), and \( F \), respectively.
- The incenter \( I \) is the intersection of the angle bisectors of \( \triangle ABC \).
- The line through \( I \) intersects \( AD \) at \( K \) and \( EF \) at \( L \).

#### 2. Key Properties of the Incircle and Incenter:
- The incenter \( I \) is equidistant from all three sides of the triangle.
- The points \( D \), \( E \), and \( F \) are the points of tangency, and the segments \( ID \), \( IE \), and \( IF \) are perpendicular to the respective sides \( BC \), \( CA \), and \( AB \).
- The line \( AD \) is the angle bisector of \( \angle BAC \).

#### 3. Symmetry and Harmonic Properties:
- The line \( EF \) is known as the Gergonne line or Soddy line of the triangle. It is the radical axis of the incircle and the nine-point circle of \( \triangle ABC \).
- The line \( AD \) is the angle bisector of \( \angle BAC \), and it passes through the incenter \( I \).

#### 4. Harmonic Division and Symmetry:
- The points \( E \) and \( F \) are symmetric with respect to the angle bisector \( AD \). This symmetry arises because \( E \) and \( F \) are the points where the incircle is tangent to the sides \( CA \) and \( AB \), respectively.
- The line \( EF \) is perpendicular to the angle bisector \( AD \) at the point where it intersects the incircle.

#### 5. Intersection Points \( K \) and \( L \):
- Since \( I \) is the incenter, it lies on the angle bisector \( AD \).
- The line through \( I \) intersects \( AD \) at \( K \) and \( EF \) at \( L \).
- Due to the symmetry of the configuration and the fact that \( I \) is the center of the incircle, the distances from \( I \) to the points \( K \) and \( L \) are equal. This is a consequence of the harmonic properties and the symmetry of the incircle with respect to the angle bisectors.

#### 6. Conclusion:
- By the properties of the incircle and the symmetry of the configuration, the distances \( IK \) and \( IL \) are equal.
- Therefore, we have \( IK = IL \).

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Final Answer:

\[
\boxed{IK = IL}
\]