Given $\mathrm{ABCD}$ is a rhombus and diagonals intersect at $\mathrm{E}$
Since in a rhombus diagonals bisect each other

but since they are opposite to each other they are of opposite signs $\Rightarrow \overrightarrow{\mathrm{EA}}=-\overrightarrow{\mathrm{EC}}$ and $\overrightarrow{\mathrm{EB}}=-\overrightarrow{\mathrm{ED}}$
$\Rightarrow \overrightarrow{\mathrm{EA}}+\overrightarrow{\mathrm{EC}}=\overrightarrow{\mathrm{O}}$
...(i) and $\overrightarrow{\mathrm{EB}}+\overrightarrow{\mathrm{ED}}=\overrightarrow{\mathrm{O}}$
Add (1) and (2)
$\Rightarrow \overrightarrow{\mathrm{EA}}+\overrightarrow{\mathrm{EB}}+\overrightarrow{\mathrm{EC}}+\overrightarrow{\mathrm{ED}}=\overrightarrow{\mathrm{O}}$