In a quadrilateral \(A B C D, M\) and \(N\) are the mid-points of the sides \(A B\) and \(C D\) respectively. If \(\overline{A D}+\overline{B C}=t \overline{M N}\), then \(t=\)

\((\mathrm{C})\)
Let \(\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{m}, \bar{n}\) be the position vectors of \(A, B, C\), \(\mathrm{D}, \mathrm{M}, \mathrm{N}\) respectively. \(\mathrm{M}\) and \(\mathrm{N}\) are the midpoints of \(\mathrm{AB}\) and CD respectively.
\(\overline{\mathrm{m}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}\), and \(\overline{\mathrm{n}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{d}}}{2} \Rightarrow \overline{\mathrm{a}}+\overline{\mathrm{b}}=2 \overline{\mathrm{m}}\) and \(\overline{\mathrm{c}}+\overline{\mathrm{d}}=2 \overline{\mathrm{n}}\)
We have \(\overline{\mathrm{AD}}+\overline{\mathrm{BC}}=\mathrm{t}(\overline{\mathrm{MN}})\)
\((\bar{d}-\bar{a})+(\bar{c}-\bar{b})=t(\bar{n}-\bar{m})\)
\(\overline{\mathrm{d}}-\overline{\mathrm{a}}+\overline{\mathrm{c}}-\overline{\mathrm{b}}=\mathrm{t}(\overline{\mathrm{n}}-\mathrm{m})\)
\((\overline{\mathrm{d}}+\overline{\mathrm{c}})-(\overline{\mathrm{a}}+\overline{\mathrm{b}})=\mathrm{t}(\overline{\mathrm{n}}-\overline{\mathrm{m}})\)
\(2 \bar{n}-2 \bar{m}=t(\bar{n}-\bar{m}) \Rightarrow t=2\)