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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 the radical axis of the incircle and the circumcircle of \( \triangle DEF \), which is known as the Gergonne point configuration.

#### 3. Symmetry and Harmonic Properties:
- The line \( EF \) is perpendicular to the line joining the incenter \( I \) and the Gergonne point \( G \) 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 \( K \) and \( L \) lie on a line passing through the incenter \( I \).
- Since \( I \) is the incenter, it is symmetric with respect to the angle bisectors and the tangency points.
- The line \( EF \) is symmetric with respect to the incenter \( I \) because \( I \) is equidistant from \( E \) and \( F \).

#### 5. Using the Radical Axis Theorem:
- The line \( EF \) is the radical axis of the incircle and the degenerate circle at \( A \) (since \( A \) is the exsimilicenter of the incircle and the excircle opposite \( A \)).
- The radical axis \( EF \) is perpendicular to the line joining the centers of these circles, which is the line \( AI \).

#### 6. Conclusion Using Symmetry:
- Since \( I \) is the incenter, it lies on the angle bisectors and is equidistant from the sides of the triangle.
- The line through \( I \) intersects \( AD \) and \( EF \) symmetrically due to the properties of the incenter and the radical axis.
- Therefore, the distances \( IK \) and \( IL \) are equal because of the symmetry and the fact that \( I \) is the midpoint of the segment joining the points of intersection along the line through \( I \).

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Final Answer:

\[
\boxed{IK = IL}
\]