Analysis: (1) As long as QA and AP are represented by an algebraic expression with t, QA = AP is used to solve it; (2) The areas of △QAC and △APC can be calculated separately; (3) As in Example 4, the solution should be divided into two situations.

Solution: (1) For any time t, AP = 2t, DQ = t and QA = 6-t. 。

When QA = AP, △ QA=AP is an isosceles right triangle.

That is 6-t = 2t.

The solution is t = 2 (seconds).

So when t = 2 seconds, △QAP is an isosceles right triangle.

(2) high DC at the edge of △QAC, QA = 6-t, QA = 12.

∴S△QAC= QA？ DC= (6-t)？ 12=36-6t。

∫In△APC，AP = 2t，BC = 6，

∴S△APC= Associated Press? BC=？ 2t？ 6=6t。

∴S quadrilateral qapc = s △ qac+s △ APC = 36-6t+6t = 36 (cm2).

From the calculation results, it is found that the area of quadrilateral QAPC remains unchanged during the movement of point P and point Q (it can also be considered that the sum of the distances from point P and point Q to diagonal AC remains unchanged).

(3) According to the meaning of the problem, it can be divided into two situations to solve:

When, △ QAP ∽△ ABC.

∴ .

The solution is t = 1.2 (s).

∴, △ QAP ∽△ ABC at t =1.2s.

When, △ PAQ ∽△ ABC.

∴ .

The solution is t = 3 (seconds).

∴, △ PAQ ∽△ ABC at t = 3s. Hope to adopt, thank you.