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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 need to prove 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 point where the angle bisectors of \( \triangle ABC \) meet, and it is equidistant from all three sides of the triangle.
- The line through \( I \) intersects \( AD \) at \( K \) and \( EF \) at \( L \).

#### 2. Key Properties of the Incircle and Incenter:
- The points \( D \), \( E \), and \( F \) are the points of tangency of the incircle with the sides of the triangle.
- The line \( AD \) is the angle bisector of \( \angle BAC \).
- The line \( EF \) is known as the Simson line of the incenter \( I \) with respect to the triangle \( \triangle DEF \). This line is perpendicular to the line joining the circumcenter of \( \triangle DEF \) and the incenter \( I \).

#### 3. Symmetry and Harmonic Properties:
- The points \( E \) and \( F \) are symmetric with respect to the angle bisector \( AD \) because \( I \) lies on the angle bisector \( AD \).
- The line \( EF \) is the radical axis of the incircle and the nine-point circle of \( \triangle DEF \), and it has specific harmonic properties related to the incenter \( I \).

#### 4. Using the Radical Axis Theorem:
- The line \( EF \) is the radical axis of the incircle and the degenerate circle at \( I \) (a point circle). This implies that any line through \( I \) will intersect \( EF \) in a way that maintains certain symmetry properties.
- Since \( I \) is the incenter, it is equidistant from the sides of the triangle, and the line \( AD \) is an angle bisector. This symmetry ensures that the distances from \( I \) to the points of intersection \( K \) and \( L \) are equal.

#### 5. Harmonic Division and Symmetry:
- The points \( K \) and \( L \) are defined by the intersections of a line through \( I \) with \( AD \) and \( EF \), respectively.
- Due to the symmetry of the configuration and the fact that \( I \) is the incenter, the distances \( IK \) and \( IL \) are equal. This can be proven using projective geometry or by noting that the line \( EF \) is symmetric with respect to the angle bisector \( AD \).

#### 6. Conclusion:
- By the properties of the incenter, the angle bisector, and the Simson line, the distances \( IK \) and \( IL \) are equal.
- Therefore, we have \( IK = IL \).

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Final Answer:

\[
\boxed{IK = IL}
\]