In p-abcd, PA = AC = 2, Pb = PD = geng6, (2) Given that point E is on PD, PE: ed = 2:1, and point F is the midpoint of edge PC, it is proved that BF is parallel to plane AEC

Proof: connect BD, AC and O, take the midpoint g of PE, connect BG and FG,
And PE: ed = 2:1
So e is the midpoint of DG and G is the midpoint of PE
And because f is the midpoint of PC
So there is FG ‖ EC in triangular PEC
So FG ‖ plane AEC
Because ABCD is a square
So o is the midpoint of AC and E is the midpoint of DG
So there is BG ‖ OE in triangle bgd
So BG ‖ plane AEC
So plane BFG ‖ plane AEC
So BF is parallel to plane AEC

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