Question

ABCD is a rhombus, whose diagonals intersect at E. Then \(\overrightarrow{EA}+\overrightarrow{EB}+\overrightarrow{EC}+\overrightarrow{ED}\) equals to :

• A. \(\overrightarrow{0}\)
• B. \(\overrightarrow{AD}\)
• C. \(2\overrightarrow{BC}\)
• D. \(2\overrightarrow{AD}\)

Answer 1

The Correct Option is A

In a rhombus, the diagonals bisect each other at right angles and divide the rhombus into four congruent right-angled triangles. Let \(O\) be the point of intersection of the diagonals.

For a rhombus \(ABCD\) with diagonals intersecting at \(E\), the diagonals bisect each other, meaning \(\overrightarrow{EA} = -\overrightarrow{EC}\) and \(\overrightarrow{EB} = -\overrightarrow{ED}\).

Therefore, we have:

\( \overrightarrow{EA} + \overrightarrow{EB} + \overrightarrow{EC} + \overrightarrow{ED} = \overrightarrow{EA} + \overrightarrow{EB} - \overrightarrow{EA} - \overrightarrow{EB} \)

This simplifies to:

\( \overrightarrow{0} \)

The sum of the vectors is indeed \(\overrightarrow{0}\), verifying that the correct answer is \(\overrightarrow{0}\).