(1)∫D is the midpoint of BC.

∴BD=CD

∫AC∨BG

∴∠CFG=∠BGF

In △BGD and △CFD,

Large CFG = ∠ BGF

Attached ∠BDG=∠CDF

Number BD=CD

∴△BGD≌△CFD(AAS)

∴BG=CF

(2) a: be+cf < ef。

Proof: ∫△BGD?△CFD

∴CF=BG,FD=GD

∵DE⊥GF

∴∠GDE=∠FDE=90

At △EDG and △EDF,

Large FD=GD

Attached ∠GDE=∠FDE

Number ED=ED

∴△EDG≌△EDF(SAS)

∴EG=EF

∫e b+ BG > EG，CF=BG

∴EB+CF>EF

Absolutely accurate! ! ! All by yourself! ! ! Hope to adopt! ! !