Problem 1341

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 $|AD| = \sqrt{|BD| \cdot |CD|}$.