Cho đường tròn (O;R), kẻ tia Bx là tiếp tuyến tại B của đường tròn (O;R), trên tia Bx lấy điểm A cách tâm O một khoảng bằng 2R. Lấy điểm C khác B và thuộc (O) sao cho AB = AC, BC cắt OA tại H

To solve the problem step by step, let's define the points and relationships clearly.

### 1. Notations and Setup
- \( O \): the center of the circle with radius \( R \).
- \( B \): the point of tangency on the circle.
- \( A \): the point on the line \( Bx \) such that \( OA = 2R \).
- \( C \): a point on the circle such that \( AB = AC \).
- \( H \): the intersection of line \( BC \) and line \( OA \).

### a) Calculate \( \sin \angle OAB \) and determine the measure of angle \( AOB \).

#### Using Geometry:
1. **Triangle \( OAB \)**:
- Since \( AB \) is tangent to the circle at \( B \), \( \angle OBA = 90^\circ \).
- In triangle \( OAB \), we have \( OA = 2R \) and \( OB = R \).

2. **Using the Pythagorean theorem**:
\[
AB^2 + OB^2 = OA^2
\]
\[
AB^2 + R^2 = (2R)^2 \implies AB^2 + R^2 = 4R^2 \implies AB^2 = 3R^2 \implies AB = R\sqrt{3}
\]

3. **Calculating \( \sin \angle OAB \)**:
\[
\sin \angle OAB = \frac{OB}{OA} = \frac{R}{2R} = \frac{1}{2}
\]
Therefore, \( \angle OAB = 30^\circ \).

4. **Calculate \( \angle AOB \)**:
- Since \( \angle OBA = 90^\circ \) and \( \angle OAB = 30^\circ \), we have:
\[
\angle AOB = 180^\circ - \angle OAB - \angle OBA = 180^\circ - 30^\circ - 90^\circ = 60^\circ
\]

### b) Prove that \( AC \) is tangent to the circle \( (O) \).

From the property of tangents:
- Since \( AB = AC \) (given) and \( A \) is external, triangles \( OAB \) and \( OAC \) are congruent by the Side-Side-Side (SSS) postulate.
- This means that \( \angle OAB = \angle OAC \).
- Because we know \( \angle OBA = \angle OCA = 90^\circ \), this means that \( AC \) must also be tangent to the circle at \( C \).

### c) Calculate \( BC \) in terms of \( R \).

Using the Law of Cosines in triangle \( ABC \):
1. **We know**:
- \( AB = AC = R\sqrt{3} \)
- \( \angle ACB = 90^\circ - \angle OAB = 60^\circ \).

2. **Using the Law of Cosines**:
\[
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle ACB)
\]
\[
BC^2 = (R\sqrt{3})^2 + (R\sqrt{3})^2 - 2 \cdot (R\sqrt{3}) \cdot (R\sqrt{3}) \cdot \frac{1}{2}
\]
\[
BC^2 = 3R^2 + 3R^2 - 3R^2 = 3R^2
\]

3. **Thus**:
\[
BC = \sqrt{3} R
\]

In summary, we have:
- \( \sin \angle OAB = \frac{1}{2} \) and \( \angle AOB = 60^\circ \).
- \( AC \) is proven to be tangent to the circle.
- \( BC = \sqrt{3} R \).