When AE is translated to BCFC 1, E and F coincide, A is located in the center of ABA 1B 1, and Let be be the point G, that is, find the cosine of the middle angle D 1FG of triangle D 1FG, so that the side length is 1. D1f = 5 (0.5)/2 = gfgd1= {0.52+[2 (0.5)/2] 2} (0.5) = 6 (0.5)/2 Reuse Cosd 1fg.