Triangle $ ABC$ is scalene with angle $ A$ having a measure greater than 90 degrees. Determine
the set of points $ D$ that lie on the extended line $ BC$, for which
\[ |AD|\equal{}\sqrt{|BD| \cdot |CD|}\]
where $ |BD|$ refers to the (positive) distance between $ B$ and $ D$.
1. **Understanding the Problem:**
   We need to find the set of points \( D \) on the extended line \( BC \) such that \( |AD| = \sqrt{|BD| \cdot |CD|} \). This condition is reminiscent of the power of a point theorem, which states that for a point \( D \) relative to a circle, the product of the lengths of the segments of any line through \( D \) that intersects the circle is constant.

2. **Using the Power of a Point Theorem:**
   The given condition \( |AD|^2 = |BD| \cdot |CD| \) suggests that \( D \) lies on the radical axis of the circle passing through \( A, B, \) and \( C \). This circle is the circumcircle of \( \triangle ABC \).

3. **Constructing the Circumcircle:**
   To find the circumcircle of \( \triangle ABC \), we need to find the perpendicular bisectors of any two sides of \( \triangle ABC \). The intersection of these bisectors is the circumcenter \( O \) of the triangle.

4. **Locating Point \( D \):**
   Since \( D \) must satisfy the power of a point condition relative to the circumcircle, \( D \) must lie on the line \( BC \) extended. The power of point theorem tells us that the power of \( D \) with respect to the circumcircle is \( |DA|^2 = |BD| \cdot |CD| \).

5. **Special Case Analysis:**
   - If \( A \) is the circumcenter, then \( A \) is equidistant from \( B \) and \( C \), and \( \angle A \) cannot be greater than \( 90^\circ \) in a scalene triangle.
   - Since \( \angle A \) is greater than \( 90^\circ \), \( A \) is not the circumcenter, and the circumcenter \( O \) lies outside \( \triangle ABC \).

6. **Conclusion:**
   The point \( D \) must lie on the line \( BC \) such that the power of \( D \) with respect to the circumcircle is equal to \( |AD|^2 \). This implies that \( D \) is the point where the perpendicular from \( A \) to \( BC \) (or its extension) intersects \( BC \). However, since \( \angle A \) is obtuse, the perpendicular from \( A \) to \( BC \) will not intersect \( BC \) within the segment \( BC \), but rather on its extension.

The final answer is \( \boxed{ D } \) lies on the extension of \( BC \) such that \( |AD| = \sqrt{|BD| \cdot |CD|} \).