( 1) AP = 4x，QC = 3x。 AB = BC = 20, AC = 30, so: AQ=30-3x.

When PQ‖BC, there is AP/AB=AQ/AC.

Namely: 4x/20=(30-3x)/30.

Solution: x= 10/3

That is, when the exercise time is 10/3 seconds, PQ‖BC.

(2) according to S △ bcq: S △ ABC = 1: 3, QC=( 1/3)AC= 10, S△BCQ=( 1/3)S△ABC.

So the movement time of q and p is 10/3.

It can be seen from (1) that PQ and BC are exactly parallel at this time.

So: △ABC∽△APQ

So: s △ apq/s △ ABC = (20/30) 2 = 4/9.

Namely: S△APQ=(4/9)S△ABC.

So: s △ pbq = s △ ABC-s △ apq-s △ bqc = s △ ABC-(4/9) s △ ABC-(1/3) s △ ABC = (2/9) s △ ABC.

So: s △ bpq: s △ ABC = 2: 9.

(3) Yes.

When △APQ∽△CQB, there is AP/CQ=AQ/BC=4/3.

Because BC=20,

So: you can get AQ=80/3.

So: QC=30-(80/3)= 10/3.

So the time for P and Q to move is (10/3)/3= 10/9.

So: AP=4*( 10/9)=40/9.

That is, the length of AP is 40/9 cm.

( 1)ED=DA，ea = EB = EC。

Prove:

∵CE⊥BD，

△ ced is a right triangle.

∫∠BDC = 60，

∴∠ECD=30。

∴CD=2DE.

CD = 2DA，

∴DE=DA.

(2) Yes, △ADE∽△AEC

Prove:

∵CE⊥BD,∠BDC=60

∴∠ACE=30，∠ADE= 120

DE = DA

∴∠DAE=∠DEA=30

∴∠DAE=∠DEA=∠ACE

∴△AEC∽△ADE (three angles are equal)

(3)ED = DA，EA=EB=EC

If the height of △AEB is h, then H = aesin30 = AE/2 = Ce/2.

∴s△bec/s△bea=eb*ce/(eb*h)=ce/(ce/2)=2