If a, b, c, d are the position vectors of points A, B, C and D respectively such that
`(a-d).(b-c) = (b-d).(c-a) = 0`
then D is the

Options

A. centroid of `Delta ABC`
B. circumcentre of `DeltaABC`
C. orthocentre of `DeltaABC`
D. none of these

Answer

To solve the problem, we need to analyze the given conditions involving the position vectors of points A, B, C, and D.

### Step-by-step Solution:

1. **Understanding the Given Conditions**:
We have the equations:
\[ (a - d) \cdot (b - c) = 0 \]
\[ (b - d) \cdot (c - a) = 0 \]
These equations imply that the vectors \( (a - d) \) and \( (b - c) \) are perpendicular, and the vectors \( (b - d) \) and \( (c - a) \) are also perpendicular.

2. **Interpreting the First Condition**:
From the first condition \( (a - d) \cdot (b - c) = 0 \):
- This means that vector \( D \) lies on the line perpendicular to vector \( (b - c) \) at point \( A \).
- Therefore, \( D \) is positioned such that the line segment \( AD \) is perpendicular to the line segment \( BC \).

3. **Interpreting the Second Condition**:
From the second condition \( (b - d) \cdot (c - a) = 0 \):
- This means that vector \( D \) lies on the line perpendicular to vector \( (c - a) \) at point \( B \).
- Hence, \( D \) is positioned such that the line segment \( BD \) is perpendicular to the line segment \( AC \).

4. **Conclusion about Point D**:
Since \( D \) is the point where the altitudes from vertices \( A \) and \( B \) of triangle \( ABC \) intersect, we can conclude that:
- Point \( D \) is the orthocenter of triangle \( ABC \).

### Final Answer:
Thus, \( D \) is the orthocenter of triangle \( ABC \).

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