Let the tangent line passing through a point $A$  outside the circle with center $O$ touches the circle at $B$ and $C$. Let $[BD]$ be the diameter of the circle. Let the lines $CD$ and $AB$ meet at $E$. If the lines $AD$ and $OE$ meet at $F$, find $|AF|/|FD|$.
1. **Identify the given elements and relationships:**
   - Let the tangent line passing through a point \( A \) outside the circle with center \( O \) touch the circle at points \( B \) and \( C \).
   - Let \( [BD] \) be the diameter of the circle.
   - Let the lines \( CD \) and \( AB \) meet at \( E \).
   - Let the lines \( AD \) and \( OE \) meet at \( F \).

2. **Establish the similarity of triangles:**
   - Since \( BD \) is the diameter, \( \angle BDC = 90^\circ \).
   - \( \angle BOC = 180^\circ - 2 \times \angle BAC \) because \( \angle BAC \) is the angle subtended by the tangent at \( A \).
   - Therefore, \( \angle BDC = \angle BOC / 2 = \angle BOA \).

3. **Use the similarity of triangles:**
   - The right triangles \( \triangle ABO \) and \( \triangle EBD \) are similar because \( \angle ABO = \angle EBD \) and \( \angle BOA = \angle BDC \).
   - This implies \( AO \parallel ED \).

4. **Establish the equality of segments:**
   - Since \( AO \parallel ED \), we have \( EA = AB \).

5. **Analyze the medians in \( \triangle EBD \):**
   - In \( \triangle EBD \), \( EO \) and \( DA \) are medians.
   - By the properties of medians in a triangle, the centroid divides each median in the ratio \( 2:1 \).

6. **Determine the ratio \( EF:FO \):**
   - Since \( EF \) and \( FO \) are parts of the median \( EO \), we have \( EF:FO = 2:1 \).

7. **Set up the perpendicular \( FK \) from \( F \) to \( EB \):**
   - Let \( FK \) be the perpendicular from \( F \) to \( EB \).
   - Since \( \triangle ABO \) and \( \triangle EBD \) are similar, \( EK:KB = 2:1 \).

8. **Express \( AK \) and \( AB \) in terms of \( x \) and \( y \):**
   - Let \( AK = y \) and \( AB = x \).
   - From the similarity and the ratio \( (x+y)/(x-y) = 2 \), we solve for \( y \):
     \[
     \frac{x + y}{x - y} = 2 \implies x + y = 2(x - y) \implies x + y = 2x - 2y \implies 3y = x \implies y = \frac{x}{3}
     \]

9. **Calculate the ratio \( AF:FD \):**
   - Since \( AF = AK \) and \( FD = KB \), we have:
     \[
     \frac{AF}{FD} = \frac{AK}{KB} = \frac{y}{x - y} = \frac{\frac{x}{3}}{x - \frac{x}{3}} = \frac{\frac{x}{3}}{\frac{2x}{3}} = \frac{1}{2}
     \]

The final answer is \(\boxed{\frac{1}{2}}\).