In the tetrahedron $ABCD$, faces $ABC$ and $BCD$ are both equilateral triangles with side length 2, and $AD=2\sqrt{2}$. Points $M$ and $N$ are the midpoints of edges $AB$ and $CD$, respectively. The shortest distance between $M$ and $N$ on the tetrahedron is
(A) 2
(B) $\frac{3}{2}$
(C) 1
(D) $\frac{5}{2}$
3. $(A)$.

Unfold the solid figure into a plane figure and then calculate.