在矩形ABCD中,AB=4,AD=5,P是射线BC上的一个动点,作PE⊥AP,PE交射线DC于点E,射线AE交射线BC于

（1）①△ABP∽△PCE
∴AB:BP = PC:CE => 4/x = (5-x)/CE
=> CE = 1/4*x(5-x)
②又有相似
所以 CF:BF = CE:BA => 3/8 = CE/4 => CE = 1.5
(2) tan∠PAE= 1/2 => PE = 1/2PA = > AP:PE = 2 = AB:PC = 4/PC =>PC =2 => BP =3