According to the parallelogram, (1)DC∨AB is obtained, and ∠2=∠FEC is deduced, and ∠ 1=∠FEC=∠2 is folded to get the answer.

(2) calculate eg = b ′ g, deduce ∠DEG=∠EGF, calculate ∠ b ′ fg = ∠ EGF through folding, and calculate de = b ′ f, and prove △ deg △ b ′ fg.

Answer:

Prove:

(1)∵ In the parallelogram ABCD, DC∨AB,

∴∠2=∠FEC，

By folding: ∠ 1=∠FEC,

∴∠ 1=∠2;

(2)∵∠ 1=∠2,

∴EG=GF，

∫AB∨DC，

∴∠DEG=∠EGF，

By folding: EC '∨B' f,

∴∠b′fg=∠egf，

∫DE = BF = B′F，

∴de=b′f，

∴△deg≌△b′fg，

∴dg=b′g.