( 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