<image>As shown in the figure, in rectangle $$ABCD$$, $$AE⊥BC$$, $$CF⊥AD$$, with the feet of the perpendiculars at $$E$$ and $$F$$, respectively. $$AE$$ and $$CF$$ intersect $$BD$$ at points $$G$$ and $$H$$, respectively, and $$AB=2 \sqrt{5}$$. 
 $$(1)$$ If $$AE =2BE$$, find the length of $$CF$$; $$(2)$$ Prove: $$BG=DH$$.