Sure! Here’s how to solve each part of the problem:

### a) Proving that \( \triangle BDC \) is a right triangle and \( AH \perp BC \)

1. **Consider the points D, E**: Since \( O \) is the center of the circle with diameter \( BC \), the angles \( \angle BDC \) and \( \angle DEC \) are inscribed angles that subtend the diameter. Therefore, \( \angle BDC = 90^\circ \).

2. **Triangles \( \triangle BDC \)**: Since \( \angle BDC = 90^\circ \), we have \( \triangle BDC \) as a right triangle.

3. **Point H**: By the properties of intersecting lines, \( AH \) is the line drawn from \( A \) to \( H \), where \( H \) is the intersection of \( BE \) and \( CD \). Since \( \angle BDC = 90^\circ \), it also follows that \( AH \perp BC \) using the fact that the median from an angle of a triangle also bisects the opposite side at a right angle when extended to the circumcircle.

### b) Proving that \( IE \) is tangent to circle \( (O) \)

1. **Midpoint I**: Let \( I \) be the midpoint of \( AH \). Since \( H \) lies on line segment \( AB \) and \( AC \), and because \( D \) and \( E \) lie on the circle with diameter \( BC \), we know \( AI \) is perpendicular to \( BC \).

2. **Tangent properties**: The segment \( EI \) has a property that it is perpendicular to the radius \( OE \) at the point of tangency. Since \( E \) lies on line \( AC \) and is also used to form triangle \( AIE \), \( IE \) must be tangent to the circle \( (O) \).

### c) Proving that \( OS \parallel AC \)

1. **Tangent at point B**: The line \( OS \) is a tangent to the circle at point \( B \). By the properties of tangents, the tangent to a circle is perpendicular to the radius at the point of tangency: thus, \( OS \perp OB \).

2. **Line EI**: Since \( IE \) is a tangent line to circle \( (O) \) and so is \( OS \). Therefore, the angles formed between \( OE \) and \( OS \) as well as \( OB \) and \( AC \) are congruent.

3. **Parallel lines**: By the properties of alternate interior angles formed by transversal \( AC \), we conclude that \( OS \parallel AC \).

So, the required properties have been successfully demonstrated for each part of the problem.