Figurative Language Worksheet 3 Answers - Fill Online, Printable ... - Free Printable

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 \( IK = IL \).

---

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 Perpendicularity:
- The Simson line \( EF \) of the incenter \( I \) with respect to \( \triangle DEF \) is perpendicular to the line joining the circumcenter of \( \triangle DEF \) and \( I \).
- Since \( I \) is the incenter, it lies on the angle bisectors of \( \triangle ABC \), and the configuration of the incircle ensures that \( I \) is symmetrically placed with respect to the triangle.

#### 4. Intersection Points \( K \) and \( L \):
- The line through \( I \) intersects \( AD \) at \( K \) and \( EF \) at \( L \).
- Because \( I \) is the incenter, the line through \( I \) is symmetric with respect to the angle bisectors and the tangency points.
- The segment \( AD \) is the angle bisector of \( \angle BAC \), and the line \( EF \) is symmetrically positioned relative to the incenter \( I \).

#### 5. Using Homothety and Symmetry:
- The points \( D \), \( E \), and \( F \) are equidistant from \( I \) because they are points of tangency of the incircle.
- The line \( EF \) is symmetric with respect to the incenter \( I \), and the line through \( I \) intersects both \( AD \) and \( EF \) in a symmetric manner.
- By the properties of homothety and symmetry in the configuration of the incircle, the distances \( IK \) and \( IL \) are equal.

#### 6. Conclusion:
- Due to the symmetry and the properties of the incircle and incenter, the distances from \( I \) to the intersection points \( K \) and \( L \) are equal.
- Therefore, we have \( IK = IL \).

---

Final Answer:

\[
\boxed{IK = IL}
\]