It is known that the bottom surface of the pyramid p-abcd is a right angled trapezoid, ab ‖ DC, ∠ DAB = 90 °. Please see the following It is known that the bottom of p-abcd is a right angled trapezoid, ab ‖ DC, ∠ DAB = 90 °, PA ⊥ bottom ABCD, PA = ad = DC = 1 / 2, ab = 1, e and m are the midpoint of edge PD and PC respectively 1. Verification: AE ⊥ PCD 2. Find a point n on the line AB so that the PDA of Mn ‖ plane

1. ∵ PA ⊥ surface ABCD
∴PA⊥CD
And ∠ DAB = 90 ° ab ‖ DC
∴CD⊥AD
⊥ CD ⊥ pad
Also AE ∈ pad
∴CD⊥AE
And △ pad is an isosceles right triangle, and point E is the midpoint of PD
∴AE⊥PD
⊥ AE ⊥ PCD
2. If an = 1 / 4, then Mn ‖ plane PDA
∵ EM is the median of △ PCD
The results show that EM = DC / 2 = 1 / 4, and EM ‖ DC ‖ an
The quadrangle an me is a parallelogram (it is not necessary to prove that it is a rectangle)
∴MN∥AE
The PDA of ∥ Mn ∥ plane

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