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The problem you provided involves geometry with circles and tangents, and is written in Vietnamese. I'll translate and solve each part of the problem.

**Problem Translation and Solution:**

**Problem 4**: Two distinct points \( B \) and \( C \) lie on a circle \( (O) \), and segment \( BC \) does not pass through the center \( O \). Tangents from \( B \) and \( C \) to the circle intersect at point \( A \).

**a)** Prove that \( AO \) is the perpendicular bisector of segment \( BC \).
**b)** Draw the perpendicular diameter \( CD \) of circle \( (O) \). Prove that \( BD \) is parallel to \( OA \).

### Solution:

**a) Proving \( AO \) is the perpendicular bisector of \( BC \)**:
- In circle geometry, the tangent at any point on the circle is perpendicular to the radius at the point of tangency. Therefore, \( AB \) is perpendicular to \( OB \) and \( AC \) is perpendicular to \( OC \).
- Since \( OB \) and \( OC \) are radii of the circle and \( B \) and \( C \) lie on the circle, \( OB = OC \).
- \( AB \) and \( AC \) are tangents from point \( A \) to the circle, hence \( AB = AC \) (tangents from a common external point to a circle are equal).
- Considering triangle \( AOB \) and triangle \( AOC \), we observe:
  - \( OB = OC \) (radii),
  - \( AB = AC \) (equal tangents),
  - \( OA \) is common to both triangles.
- By the Side-Side-Side (SSS) criterion, triangles \( AOB \) and \( AOC \) are congruent.
- Consequently, \( \angle AOB = \angle AOC \), indicating \( AO \) bisects \( \angle BOC \).
- Since \( \angle BOC \) is the angle subtended by \( BC \) at the circle’s center, and \( \angle BOC \) is bisected by \( AO \), \( AO \) is perpendicular to \( BC \) (as \( \angle BOC \) is twice the angle subtended by \( BC \) on the circle's circumference).
- Therefore, \( AO \) is the perpendicular bisector of \( BC \).

**b) Proving \( BD \) is parallel to \( OA \)**:
- After drawing diameter \( CD \) which is perpendicular to \( BC \) (by definition of a diameter being a special chord that passes through the center and splits the circle into two equal halves), \( CD \) aligns with \( OA \) because \( AO \) is the perpendicular bisector of \( BC \) (from part a).
- \( BD \) is a line segment from the circumference of the circle to \( D \), the endpoint of the diameter.
- Since \( CD \) (or \( OD \) as part of the diameter) is vertical to \( BC \), and \( AO \) is vertical to \( BC \), \( BD \) and \( OA \) are parallel because both are perpendicular to the same line \( BC \).

Would you like further details on any part of this solution, or do you have any other questions on this topic?

### Related Questions:
1. What would happen if point \( A \) was inside the circle?
2. How can the concept of congruent triangles be used to solve other circle geometry problems?
3. How do the properties of a circle’s diameter help in proving theorems about angles in a circle?
4. Could you find the area of triangle \( AOB \) if the radius of the circle and the length \( AB \) were known?
5. What other geometric figures could you investigate using similar concepts of tangents and diameters?

### Tip:
When working with circle geometry, always consider properties like equal tangents from a point outside the circle and perpendicularity between the tangent and the radius at the point of tangency. These often provide critical insights into proving various geometrical theorems.

Given two distinct points B and C on a circle (O), with BC not passing through the center O. Tangents from B and C intersect at A. a) Prove AO is the perpendicular bisector of BC. b) Draw the diameter CD and prove BD is parallel to OA.

Learn how to solve a geometric problem involving tangents, circles, and perpendicular bisectors. We explore the relationship between tangents and the radius, and prove AO bisects BC and BD is parallel to OA, using key theorems such as the Congruent Triangles Theorem and properties of tangents.

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