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.