Given:
- AB is a diameter of a circle with center O.
- AC is a chord of the circle.
- Angle BAC = 30 degrees.
- The tangent at C intersects AB at point D.

To prove:
BC = BD

Proof:

1. Angle in a semicircle:
In a circle, if a chord is drawn with one endpoint on the circumference and the other endpoint on the diameter, the angle formed at the circumference is always a right angle (90 degrees).
Therefore, angle BOC = 90 degrees.

2. Tangent and radius:
A tangent to a circle is perpendicular to the radius at the point of tangency.
Therefore, CD is perpendicular to AC at point C.

3. Triangle ACD:
In triangle ACD, angle CDA = 90 degrees (perpendicularity of CD).
Since angle BAC = 30 degrees, angle CAD = (180 - 90 - 30) degrees = 60 degrees.
Therefore, triangle ACD is a 30-60-90 triangle.

4. Properties of a 30-60-90 triangle:
In a 30-60-90 triangle, the side opposite the 30-degree angle is half the length of the hypotenuse, and the side opposite the 60-degree angle is √3 times the length of the side opposite the 30-degree angle.
In triangle ACD, AD is the side opposite the 30-degree angle, AC is the hypotenuse, and CD is the side opposite the 60-degree angle.
Therefore, AD = AC/2 and CD = (√3)AD.

5. Triangle CBD:
In triangle CBD, angle CBD = angle CBA + angle ABD = 90 degrees + 30 degrees = 120 degrees.
Therefore, triangle CBD is a 30-60-90 triangle.

6. Properties of triangle CBD:
In triangle CBD, BD is the side opposite the 30-degree angle, BC is the hypotenuse, and CD is the side opposite the 60-degree angle.
Since triangle CBD is a 30-60-90 triangle, BD = BC/2 and CD = (√3)BD.

7. Conclusion:
From step 4, we know that CD = (√3)AD. Substituting AD = AC/2 from step 3, we have CD = (√3)(AC/2) = (√3/2)AC.
From step 6, we know that CD = (√3)BD. Equating the expressions for CD, we have (√3/2)AC = (√3)BD.
Dividing both sides by (√3), we get AC/2 = BD.
Therefore, BC = BD (subtracting AC/2 from both sides).

Hence, we have proved that BC = BD.