If ABCD is a rhombus whose diagonals intersect at E, then `vec(EA) + vec(EB) + vec(EC) + vec( ED)` equals

Options

A. `vec(0)`
B. `vec(AD)`
C. `2 vec(BC)`
D. `2 vec(AD)`

Answer

To solve the problem, we need to analyze the properties of a rhombus and the vectors associated with its diagonals. Here’s a step-by-step solution:

### Step 1: Understand the properties of a rhombus
A rhombus is a quadrilateral where all sides are of equal length, and its diagonals bisect each other at right angles. Let the vertices of the rhombus be labeled as A, B, C, and D, with the diagonals AC and BD intersecting at point E.

### Step 2: Represent the vectors
We can represent the vectors from point E to the vertices A, B, C, and D as follows:
- Let \( \vec{EA} \) be the vector from E to A.
- Let \( \vec{EB} \) be the vector from E to B.
- Let \( \vec{EC} \) be the vector from E to C.
- Let \( \vec{ED} \) be the vector from E to D.

### Step 3: Analyze the relationships between the vectors
Since E is the midpoint of both diagonals:
- The vector \( \vec{EA} \) is equal in magnitude but opposite in direction to \( \vec{EC} \). Therefore, we can write:
\[ \vec{EA} = -\vec{EC} \]

- Similarly, for the other diagonal:
\[ \vec{EB} = -\vec{ED} \]

### Step 4: Combine the vectors
Now, we can add the vectors together:
\[ \vec{EA} + \vec{EB} + \vec{EC} + \vec{ED} \]

Substituting the relationships we found:
\[ \vec{EA} + \vec{EB} + (-\vec{EA}) + (-\vec{EB}) = 0 \]

### Step 5: Conclusion
Thus, we find that:
\[ \vec{EA} + \vec{EB} + \vec{EC} + \vec{ED} = \vec{0} \]

### Final Answer
The result is the zero vector:
\[ \vec{EA} + \vec{EB} + \vec{EC} + \vec{ED} = \vec{0} \]