已知菱形ABCD的边长为5,∠DAB=60°.将菱形ABCD绕着A逆时针旋转得到菱形AEFG,设∠EAB=α,且0°<α<90°,连接DG、BE、CE、CF.
(1)如图(1),求证:△AGD≌△AEB;
(2)当α=60°时,在图(2)中画出图形并求出线段CF的长;
(3)若∠CEF=90°,在图(3)中画出图形并求出△CEF的面积.
解:(1)∵菱形ABCD绕着点A逆时针旋转得到菱形AEFG,
∴AG=AD,AE=AB,∠GAD=∠EAB=α.
∵四边形AEFG是菱形,
∴AD=AB.
∴AG=AE.
∴△AGD≌△AEB.
(2)解法一:如图(1),当α=60°时,AE与AD重合,