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Mathematics problems in the national college entrance examination II
(1) Let the midpoint of AC and BD be O.

Connecting OE

Because e is the midpoint of the hypotenuse of the right triangle pad, DE=EP.

O is the midpoint of BD, so DO=BO.

In triangular PBD, DE:DP=DO:DB, so △DEO is similar to △DPB EO∨PB.

EO belongs to planar AEC.

So Pb Σ plane AEC

(2) A is used as AF⊥PB of point F.

Because PA⊥ plane ABCD, PA⊥BC

Because ABCD is rectangular,

So BC⊥AB

So BC⊥ aircraft PAB

So BC⊥AF

Because AF⊥PB

So AF⊥ aircraft PBC

The volume of P-ABD v =1/3× s× h.

= 1/3×( 1/2×ab×AD)×PA

It is known that the length and volume substitution of PA AD can be obtained.

AB=3/2

In the right triangle PAB

1/2xpaxab =1/2xpaxaf (area formula)

PB? =PA? +AB? You can get PB= root number 13/2.

So AF=PAXAB/PB=3 times the root number 13/ 13.

Therefore, the distance from A to PBC is three times that of the root number 13/ 13.

Please accept questions that pure hands don't understand.