It is known that the parallelepiped abcd-a'b'c'd ', e, F, G and H are the midpoint of edges a'd', d'c ', c'c and ab respectively. It is proved that e, F, G, h are coplanar

It is known that the parallelepiped abcd-a'b'c'd ', e, F, G and H are the midpoint of edges a'd', d'c ', c'c and ab respectively. It is proved that e, F, G, h are coplanar

Connect a'B, BC ', a'c',
Connect AC, CD ', ad'
Yi Zheng: AC / / a'c ', a'B / / d'c
Know plane a'bc '/ / ACD' (if two intersecting lines on one plane are parallel to two intersecting lines on the other plane, then the two planes are parallel to each other)
Another easy syndrome: EF / / a'c '/ / AC, FG / / a'B / / d'c. knowing plane EFG / / plane a'bc' / / ACD '
It is also known that the plane EFG is equidistant from the other two planes
Thus, let m and n be points on plane a'bc 'and ACD' respectively, then: plane EFG bisects line Mn
A and B are such two points, so their midpoint h must be on the plane EFG
E, F, G and H are coplanar

In the parallelepiped abcd-a1b1c1d1, let the vector AC1 be equal to the vector ab of x times plus the vector BC of 2Y times plus the vector C1C of 3Z times
So what is x + y + Z?

Vector AC1 is equal to x times vector AB plus 2Y times vector BC plus 3Z times vector C1C
And vector AC1 = vector AB + vector BC1,
Vector BC1 = vector BC + vector CC1, vector CC1 = - vector C1C,
Then, vector AC1 = vector AB + vector BC - vector C1C,
That is to say,
X=1,2Y=1,3Z=-1.
X=1,Y=1/2,Z=-1/3,
X+Y+Z=(1+1/2-1/3)=(6+3-2)/6=7/6.

In the parallelepiped abcd-a1b1c1d1, try to find x AC + Ab1 + AD1 = xac1 (all vectors divide by x)

AC=AB+AD
AB1=AA1+AB
AD1=AD+AA1
So AC + Ab1 + AD1 = 2 (AB + AD + Aa1) = 2ac1
And because AC + Ab1 + AD1 = xac1
So x = 2

In the cube abcd-a1b1c1d1, e and F are the midpoint of CD and BC respectively. The cosine value of the angle formed by AE and C1F is calculated

Connect a to the midpoint g, Ge, Gd of a1d1
∵ CD ⊥ aa1dd1
∴DE⊥DG AG∥C1F
The angle of AE and C1F is gae
Let the side length of cube be 2
Then Ag = AE = √ 5 eg = √ [(√ 5) ^ 2 + 1] = √ 6
∴cos∠GAE=[(√5)^2+(√5)^2-(√6)^2]/2√5*√5=2/5