As shown in Figure (2), △DEF starts from the position in Figure (1) and moves uniformly along CB to △ABC at the speed of 1 cm/s, while △DEF moves, point P starts from the vertex B of △ABC and moves uniformly along BA to point A at the speed of 2 cm/s. When the vertex D of △DEF moves to AC,
(1) When what is the value of t, is point A on the vertical line of PQ?
(2) Connect PE, let the area of quadrilateral APEC be y(cm2), and find the functional relationship between y and t; Is there a moment when t minimizes the area y? If it exists, find the minimum value of y; If it does not exist, explain why.
(3) Is there a moment t that makes P, Q and F in the same straight line? If it exists, find the value of t at this time; If it does not exist, explain why. Solution: (1)∵ Point A is on the vertical line of PQ,
∴AP = AQ。
∠∠DEF = 45，∠ACB = 90，∠DEF+∠ACB+∠EQC = 180，
∴∠EQC = 45。
∴∠DEF =∠EQC。
∴CE = Chongqing.
From the meaning of the question: CE = t, BP =2 t,
∴CQ = t
∴AQ = 8 tons
In Rt△ABC, we can know from Pythagorean theorem that AB = 10 cm.
Then AP =10-2 t.
∴ 10-2 t = 8-t
Solution: t = 2.
Answer: When t = 2 s, point A is on the vertical line of PQ line.
(2) Make PM⊥BE and ∴ ∠ cross into m, BMP = 90.
In Rt△ABC and Rt△BPM,
∵BC = 6 cm，CE = t，∴ BE = 6-t
When t = 3, y is minimum = 84/5a; when t = 3s, the area of quadrilateral APEC is minimum, with the minimum area of 84/5cm2.
(3) Suppose there is a certain moment t, so that points P, Q and F are on the same straight line.
Through p, through AC to n,
∫∠ACB = 90, b, C(E) and f are on the same straight line,
∴∠QCF = 90，∠QCF = ∠PNQ。
∠∠FQC =∠PQN，
∴△QCF∽△QNP。
Solution: t = 1.
A: When t = 1s, P, Q and F are on the same straight line.
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