延长AB,作BE=DQ,连接CE 则△CDQ≌△CBE ∴∠DCQ=∠BCE,DQ=BE,CQ=CE ∴∠QCE=∠BCE+∠BCQ=∠DCQ+∠BCQ=90º 设DQ=X,BP=Y 则AQ=1-X,AP=1-Y PE=DQ+PB=X+Y,PQ=△APQ周长-AQ-AP=2-(1-X)-(1-Y)=X+Y 则△QCP≌△ECP (...