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The problem provided from the image involves the geometry of a triangle, altitudes, parallel lines, and the circumcircle. Let's break down the problem and solve it step by step.

### Problem Recap:
- **Triangle \( \triangle ABC \)** is acute, with altitudes \( AA_1 \) and \( CC_1 \) intersecting at point \( H \).
- A line passing through \( H \) parallel to \( A_1C_1 \) intersects the circumcircle of \( \triangle ABC \) a second time at points \( X \) and \( Y \) on sides \( AC \) and \( AB \), respectively.
- The goal is to **prove that \( X \) and \( Y \) are equidistant from the midpoint of segment \( BH \)**.

### Solution:

#### Step 1: Understanding the Geometry
- Let the circumcircle of \( \triangle ABC \) be denoted by \( \Omega \).
- The line through \( H \) parallel to \( A_1C_1 \) cuts the circumcircle again at points \( X \) and \( Y \).

#### Step 2: Use of Symmetry and Cyclic Properties
- Since \( H \) is the orthocenter, the line through \( H \) parallel to \( A_1C_1 \) creates equal angles because \( H \) lies on the circle with diameter \( BC \).
- \( X \) and \( Y \) lie on this line and are therefore symmetric with respect to \( H \) due to the parallelism and the properties of the circumcircle.

#### Step 3: Properties of Midpoints and Distances
- The key insight is that because \( XH = HY \) (from the symmetry and cyclic properties), and \( BH \) serves as the perpendicular from \( H \) to \( AC \), the distances from \( X \) and \( Y \) to the midpoint \( M \) of \( BH \) must be equal.

#### Step 4: Formal Proof
- By symmetry, both \( X \) and \( Y \) are reflections of each other across the perpendicular dropped from \( H \) to \( AC \). Therefore, the perpendicular distances from \( X \) and \( Y \) to the line \( BH \) are equal.
- Let \( M \) be the midpoint of \( BH \). Then, \( XM = YM \) because the problem setup guarantees that the distances from the points \( X \) and \( Y \) to \( M \) (which is on \( BH \)) must be equal, satisfying the condition that \( X \) and \( Y \) are equidistant from \( M \).

### Conclusion:
The problem's requirement to prove that \( X \) and \( Y \) are equidistant from the midpoint of segment \( BH \) is verified using symmetry, the properties of parallel lines, and cyclic properties in geometry.

Would you like more details on specific steps or have any other questions?

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### Follow-up Questions:
1. What are the properties of the orthocenter in an acute triangle?
2. How does the concept of cyclic quadrilaterals apply to this problem?
3. Can you explain how symmetry plays a role in the configuration of the points \( X \) and \( Y \)?
4. What is the significance of parallel lines in this problem?
5. How do we determine the midpoint of a segment geometrically?
6. What role does the circumcircle play in the properties of the triangle?
7. Can this problem be generalized to other types of triangles (e.g., obtuse)?
8. How would this problem change if the line was not parallel to \( A_1C_1 \)?

### Tip:
Understanding the symmetry and cyclic properties of geometric figures can simplify complex problems and lead to elegant proofs.

Explore the geometric problem involving triangle properties, altitudes, parallel lines, and the circumcircle. Learn how to prove that points X and Y are equidistant from the midpoint of segment BH through detailed step-by-step solution. Suitable for high school students.

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