As shown in the figure, ABCD and abef are two congruent squares that are not in the same plane. The points m and N are on the diagonal lines AC and BF respectively, and cm = BN. Prove: Mn / / plane BCE

As shown in the figure, ABCD and abef are two congruent squares that are not in the same plane. The points m and N are on the diagonal lines AC and BF respectively, and cm = BN. Prove: Mn / / plane BCE

AB⊥ BC.AB In ABCD, make MP ‖ ab. P ∈ BC
Then ⊿ MPC is an isosceles right triangle, MP = MC / √ 2. MP ⊥ plane BCE
That is, the distance from m to plane BCE = MC / √ 2. Similarly, the distance from n to plane BCE = Nb / √ 2 = MC / √ 2,
Mpqn is a rectangle (q is the perpendicular foot of N in be) Mn ‖ PQ ∈ plane BCE.. ‖ Mn ‖ plane BCE

As shown in the figure, let ABCD and abef be parallelograms, and they are no longer in the same plane. M and N are points on diagonal lines AC and BF respectively,
And am: FN = AC: BF

Let MP ∥ BC intersect AB with P, even NP,
∥ AM / AC = AP / AB, MP ∥ plane BCE,
AM：FN=AC：BF,
∴AM/AC=FN/BF=AP/AB,
∥ PN ∥ AF ∥ be (abef is a parallelogram),
Ψ PN ‖ plane BCE,
Plane MNP plane BCE,
The plane BCE

As shown in the figure, the side lengths of square ABCD and abef are all 1, and the plane ABCD and plane abef are perpendicular to each other. Point m moves on AC, and point n moves on BF. If cm = BN = a (0 < a < root 2), the space rectangular coordinate system is established,
(1) Finding the length of Mn
(2) When the value of a is, the length of Mn is the smallest

Take a (0.0,0) d (1,0,0) B (0,1,0), f (0,0,1), then M (1-A / √ 2,1-a / √ 2.0), n (0,1-a / √ 2, a / √ 2,) 1 MN's length L = √ [(1-A / √ 2) & sup2; + (a / √ 2) & sup2;] = √ [(a - √ 2 / 2) & sup2; + 1 / 2] 2 (a = √ 2 / 2), l has the minimum √ (1 / 2) [

ABCD and abef are parallelograms, m and N are diagonal points AC and BF, and am: FN = AC: BF

Guo m points to do mg / / NH / / AB AC AF in H connect Hg, nm through n points, because in G AB AC ad quadrilateral ABCD is a parallelogram, and GM / / DC, so GM / DC = am / AC, that is general / AB = am / AC, the same can be HN / AB = FN / FB, because am / AC = FN / FB, so Hb / AB =