20. 如图1,四边形ABCD是正方形,点G是BC上任意一点,DE⊥AG于点E,BF⊥AG于点F.
(1)求证:$DE-BF=EF$.
(2)当点G为BC边中点时,试探究线段EF与FG之间的数量关系,并说明理由.
(3)若点G为CB延长线上一点,其余条件不变.请你在图2中画出图形,写出此时DE,BF,EF之间的数量关系.(不需要证明)

20. (1)$\because$四边形$ABCD$是正方形,$BF⊥ AG,DE⊥ AG$,
$\therefore DA=AB,∠ BAE+∠ DAE=∠ DAE+∠ ADE=90°.$
$\therefore ∠ BAF=∠ ADE.$
$\therefore △ ABF≌△ DAE.$
$\therefore BF=AE,AF=DE.$
$\therefore DE-BF=AF-AE=EF.$
(2)$EF=2FG$.理由如下：
令$AB=BC=a$.
$\because$点$G$为$BC$的中点，
$\therefore BG=\dfrac{1}{2}BC=\dfrac{1}{2}a,AG=\sqrt{AB^2+BG^2}=\dfrac{\sqrt{5}}{2}a.$
$\because \dfrac{1}{2}AB× BG=\dfrac{1}{2}BF× AG,$
$\therefore BF=\dfrac{AB× BG}{AG}=\dfrac{\sqrt{5}}{5}a.$
$\therefore AF=\sqrt{AB^2-BF^2}=\dfrac{2\sqrt{5}}{5}a.$
$\therefore EF=AF-AE=AF-BF=\dfrac{\sqrt{5}}{5}a,$
$FG=AG-AF=\dfrac{\sqrt{5}}{10}a.$
$\therefore EF=2FG.$
(3)如图,$DE+BF=EF.$