Let the intersection of plane ACC 1A 1 and plane A 1BM be AE, AC∩BD=E,

The bottom quadrilateral ABCD is a parallelogram with diagonal lines bisecting each other, so e is the midpoint of BD.

If A 1C is connected in the plane ACC 1A 1 and intersects with AC 1 at n, then n is the midpoint between A 1C and AC 1.

In triangle A 1AC, an and AE are both median lines, so point m is the center of gravity of triangle ACA 1.

AM=2AE/3，

In triangle A 1DB, e is the midpoint of BD side, A 1E is the median line of BD side, M∈A 1E, AM=2AE/3,

The intersection m of AC 1 and plane BA 1D is the center of gravity of triangle ba1d.