Solution, (1) If △PAQ is an isosceles right triangle, then AQ=AP.

AQ= 10-t, AP=2t, so 2t= 10-t, t= 10/3.

(2) If you want to make △PAQ∽ABC, there are two situations.

①AQ/BC=AP/BA, that is, (10-t)/ 10=2t/20, t=5.

②AQ/ bar =AP/BC, that is, (10-t)/20=2t/ 10, t=2.

Therefore, when t=2 or t=5, △PAQ∽△ABC.

(3) Quadrilateral APCQ = △ ACP+△ ACQ = AP * BC/2+AQ * CD/2 =10t+100t =100.

That is to say, the quadrilateral APCQ is fixed at 100cm?