As shown in the figure, e is the midpoint of dd1 in cube abcb-a1b1c1d1. Try to judge the position relationship between BD1 and plane AEC, and explain the reason

As shown in the figure, e is the midpoint of dd1 in cube abcb-a1b1c1d1. Try to judge the position relationship between BD1 and plane AEC, and explain the reason


parallel.
Let f be the midpoint of AC and EF be parallel to BD1
(just look at the triangle bdc1.)
So BD1 is parallel to AEC



In the cuboid abcd-a1b1c1d1, e is the midpoint of edge BC


Connect: d1c, cross DC1 to F. know that f is the midpoint of d1c
Connect EF, which is the median line of triangle d1bc
So EF / / D1b
EF is on plane c1de
According to the theorem: BD1 / / plane c1de



P. Q and R are the midpoint of edges AB, BB1 and BC of cube abcd-a1b1c1d1 respectively


Abcd-a1b1c1d1 is a cube, ABCD is a square, AC ⊥ BD
∵ abcd-a1b1c1d1 is a cube, ∵ BB11 ⊥ planar ABCD, ∵ AC ⊥ BB1
From AC ⊥ BD, AC ⊥ BB1, we get: AC ⊥ plane bb1d, ⊥ BD1 ⊥ AC
∵ P and R are the middle points of AB and BC, ∵ PR ∥ AC, ≁ BD1 ⊥ PR, respectively
∵ abcd-a1b1c1d1 is a cube, ∵ aa1b1b is a square, ∵ Ab1 ⊥ A1B
∵ abcd-a1b1c1d1 is a cube, ∵ BC ⊥ plane aa1b1b, ∵ Ab1 ⊥ BC
From Ab1 ⊥ A1B, Ab1 ⊥ BC, Ab1 ⊥ plane a1bcd1, ⊥ BD1 ⊥ Ab1
∵ P and Q are the midpoint of AB and BB1 respectively, ∵ PQ ∥ Ab1, ∵ BD1 ⊥ PQ
From BD1 ⊥ PR and BD1 ⊥ PQ, BD1 ⊥ plane PQR is obtained



In the cube abcd-a1b1c1d1, e is the midpoint of BC. It is proved that BD1 is parallel to cide
emergency
Just a simple idea


Make DC1 midpoint m, connect me, d1c
∵ m and E are d1c and BC respectively
The me is the median of △ d1bc
∴D1B‖ME
∵ me in plane c1de
BD1 is not in plane c1de
‖ BD1 ‖ plane c1de
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