Problem 1. The altitudes AS and CJ of an acute triangle ABC intersects at
point H. The line passing through H parallel to SJ intersects a second time the
circumcircle of triangle AHJ and triangle CHS at X and Y respectively. Prove that X and Y
are equidistant from the midpoint of segment BH.

Let's break down the problem and prove that \( X \) and \( Y \) are equidistant from the midpoint of segment \( BH \).

### Problem Recap
- We have an acute triangle \( ABC \).
- Altitudes \( AS \) and \( CJ \) intersect at the orthocenter \( H \).
- A line passing through \( H \) parallel to \( SJ \) intersects the circumcircle of \( \triangle AHJ \) at point \( X \) and the circumcircle of \( \triangle CHS \) at point \( Y \).
- We need to prove that \( X \) and \( Y \) are equidistant from the midpoint of segment \( BH \).

### Step-by-Step Solution

1. **Geometry Setup:**
   - Let \( M \) be the midpoint of segment \( BH \).
   - Since \( H \) is the orthocenter, the points \( A \), \( H \), \( J \), \( S \), and \( C \) have certain properties in terms of cyclicity and perpendicularity.

2. **Parallel Line Construction:**
   - The line \( l \) through \( H \) is parallel to \( SJ \) and intersects the circumcircle of \( \triangle AHJ \) at \( X \) and the circumcircle of \( \triangle CHS \) at \( Y \).
   - The parallelism ensures that the angles subtended by \( XH \) and \( YH \) on their respective circles are equal, i.e., \( \angle XHJ = \angle YHS \).

3. **Properties of the Circumcircles:**
   - Consider the circumcircle of \( \triangle AHJ \). Since \( HJ \) is an altitude, the circumcircle of \( \triangle AHJ \) has a diameter \( AJ \) (due to the right angle at \( H \)).
   - Similarly, the circumcircle of \( \triangle CHS \) has a diameter \( CS \).

4. **Cyclic Quadrilaterals:**
   - The points \( A \), \( H \), \( J \), and \( X \) lie on a circle. 
   - The points \( C \), \( H \), \( S \), and \( Y \) lie on another circle.
   - Both \( \angle AJX \) and \( \angle CSY \) are right angles because they subtend the diameters \( AJ \) and \( CS \), respectively.

5. **Midpoint and Perpendicularity:**
   - Since \( l \parallel SJ \), the perpendicular distance from \( M \) to \( X \) and \( Y \) must be equal. 
   - Moreover, since \( M \) is the midpoint of \( BH \), \( XM \) and \( YM \) are both perpendicular to the line \( l \) (due to the symmetry about the midpoint \( M \)).

6. **Conclusion:**
   - Because the distances from \( X \) to \( M \) and \( Y \) to \( M \) along the line \( l \) are the same, \( X \) and \( Y \) are equidistant from the midpoint \( M \) of segment \( BH \).

### Final Statement
Thus, \( X \) and \( Y \) are equidistant from the midpoint \( M \) of segment \( BH \).

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Would you like more details on any specific part of the proof or further explanations?

### Related Questions

1. What is the significance of the orthocenter in an acute triangle?
2. How does the concept of cyclic quadrilaterals apply to this problem?
3. Why is the midpoint \( M \) important in proving equidistance?
4. What role do the parallel lines play in this proof?
5. Can this result be generalized to any triangle, or is it specific to acute triangles?
6. How does the circumcircle's property of right angles apply to this proof?
7. What would change if the triangle were obtuse instead of acute?
8. How would this problem change if \( X \) and \( Y \) were on different lines instead of the same one?

### Tip
In geometry problems involving circumcircles and cyclic quadrilaterals, always look for angle properties and parallel lines, as they often lead to crucial insights for solving the problem.

Explore the proof showing that points X and Y, intersections on circumcircles of triangles AHJ and CHS respectively, are equidistant from the midpoint of segment BH in an acute triangle. Understand the properties of orthocenters, circumcircles, and cyclic quadrilaterals in geometry. Suitable for advanced high school students.

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