In the known cube abcd-a1b1c1d1, we prove: (1) AC ⊥ plane b1d1db; (2) BD1 ⊥ plane Acb1

In the known cube abcd-a1b1c1d1, we prove: (1) AC ⊥ plane b1d1db; (2) BD1 ⊥ plane Acb1

It is proved that: (1) in cube abcd-a1b1c1d1, B1B ⊥ plane ABCD, AC ⊂ plane ABCD, ⊥ AC ⊥ BB1, and ⊥ AC ⊥ BD, BD ∩ B1B = B, ⊥ AC ⊥ plane b1d1db; (2) in cube abcd-a1b1c1d1, B1B ⊥ plane ABCD, AC ⊂ plane ABCD, ⊥ AC ⊥ BD1, and ⊥ AC ⊥ BD1 are flat

In the cube abcd-a'b'c'd ', e.f.g AB.BC.AA Verification: BD 'is perpendicular to the plane EFG

[method 1] take the midpoint of CC 'as h, from EF / / GH, efgh coplanar, take the midpoint of a'd' as I, the midpoint of c'd 'as J, from GH / / ij, ghji coplanar. From eg / / HJ, efghji coplanar, the plane is parallel to the triangle ad'c.bb' and perpendicular to ABCD, then AC is perpendicular to BB '

How to calculate the angle between diagonal b'd and plane a'bc 'in cube abcd-a'b'c'd'?

Because the a'c'vertical plane bdd'b ', so a'c'vertical b'd, similarly: a'B vertical plane ADB', so a'B vertical b'd, so b'd vertical plane a'c'b, so the angle is 90 degrees

In the cube abcd-a'b'c'd ', BD, BC', DC 'are three plane diagonals respectively, and a'c is a body diagonals
The second key is to prove a'c ⊥ BD a'c ⊥ plane DBC '

It is known that the cube abcd-a'b'c'd ', O is the intersection of the diagonals of the quadrilateral ABCD. Prove: c'o / / plane ab'd', a'c ⊥ plane ab'd '

Proof: (1) to prove that a line is parallel to a face, we only need to prove that the line is parallel to a line of the face, because c'o / / AO '(o' is the intersection of diagonal lines of quadrilateral a'b'c'd '), so c'o / / plane ab'd' (2) to prove that a line is perpendicular to a face, we only need to prove that the line is perpendicular to the plane

As shown in the figure, in the cube abcd-a1b1c1d1, verify: plane acc1a1 ⊥ plane a1bd

It is proved that: Aa1 ⊥ plane ABCD ⊥ BD ⊥ AC, BD ⊥ A1A, AC ∩ A1A = a ⊥ BD ⊥ plane acc1a1 and BD ⊂ plane a1bd ⊥ plane acc1a1 ⊥ plane a1bd

In the cube abcd-a'b'c'd ', it is proved that the plane acc'a' is perpendicular to the plane a'bd?

Theorem: if a plane passes through a vertical line of another plane, the plane is perpendicular to the other plane
Because BD is perpendicular to acc'a '
So a'bd is perpendicular to acc'a

As shown in the figure, in the cube abcd-a1b1c1d1, verify: plane acc1a1 ⊥ plane a1bd

It is proved that: Aa1 ⊥ plane ABCD ⊥ BD ⊥ AC, BD ⊥ A1A, AC ∩ A1A = a ⊥ BD ⊥ plane acc1a1 and BD ⊂ plane a1bd ⊥ plane acc1a1 ⊥ plane a1bd

As shown in the figure, in the cube abcd-a1b1c1d1, verify: plane acc1a1 ⊥ plane a1bd

It is proved that: Aa1 ⊥ plane ABCD ⊥ BD ⊥ AC, BD ⊥ A1A, AC ∩ A1A = a ⊥ BD ⊥ plane acc1a1 and BD ⊂ plane a1bd ⊥ plane acc1a1 ⊥ plane a1bd

If a / b = C / D = k, does a-c / B-C = a + C / B + d hold? Why

Using the substitution method, if the denominator cannot be 0, then K is not equal to 2, and 1 / K A-1 / 2 A is not 0, then k = 1, a = B, C = D can be obtained. Because of the existence of line segment, a is not equal to 0, B is not equal to 0, C is not equal to 0, D is not equal to 0
Set up