∴AE = ab = 20cm;
(2) ∵ According to the folding, AG=AB, ∠GAE=∠BAE, ∵ Point P is the midpoint of AB,?
∴ AP= AB,∴ AP= AG,?
In Rt△APG, ∠ gap = 60, ∴ ∠ EAB = 30,
In Rt△EAB, AE= AB= cm.
(3) EH⊥AD is at point H, even BF. From folding, we know that DE=BE,
AF = FG,DF=AB,GD=AB,
∴△abf?△GDF, and: △△GDF =∠CDE, GD=CD,
? ∴ Rt△GDF≌Rt△CDE
∴ DF=DE=BE,
At Rt△DCE, DC 2 +CE 2 =DE 2.
CB = 25,CD=20,20 ^ 2+ce2 =(25-CE)2,
∴ CE=4.5,BE=25-4.5=20.5,HF=20.5-4.5= 16,
At Rt△EHF,
∫EH 2+HF 2 = Fe2,20 2 + 16 2 =FE 2,?
∴ EF= = cm