9. 如图，AB//CD，直线EF与AB，CD分别交于M，N两点，过点M作MG⊥EF交CD于G点，过点G作GH平分∠MGD，∠EMB=40°. 求∠MGH的度数.

9. 解:$\because MG⊥ EF$,
$\therefore ∠ GME=90°$.
$\therefore ∠ BMG=90°-∠ EMB=90°-40°=50°$.
$\because AB// CD$,
$\therefore ∠ BMG=∠ MGN=50°$.
$\therefore ∠ MGD=130°$.
$\because GH$平分$∠ MGD$,
$\therefore ∠ MGH=\frac{1}{2}∠ MGD=65°$.

10. [新考法]如图甲，AB//CD，若E为平面内一动点（点E不在直线AD和直线CD上），连接ED，过点E作EF//CD，且点F在点E的右侧.
(1)当点E运动到如图乙所示位置时，求证：∠A - ∠DEF = ∠ADE.
(2)直接用等式表示出∠A，∠ADE，∠DEF之间存在的所有数量关系.

10. (1)证明:$\because AB// CD$,
$\therefore ∠ A=∠ ADC$.
$\because EF// CD$,
$\therefore ∠ DEF=∠ EDC$.
$\therefore ∠ A-∠ DEF=∠ ADC-∠ EDC$,
即$∠ A-∠ DEF=∠ ADE$.
(2)解:当点$E$在$AD$左侧,且在$CD$上方时,根据(1),可得$∠ A-∠ DEF=∠ ADE$;
如图,当点$E$在$AD$右侧,且在$CD$上方时,
$\because AB// CD$,
$\therefore ∠ A=∠ ADC$.
$\because EF// CD$,
$\therefore ∠ DEF=∠ EDC$.
$\therefore ∠ A+∠ ADE=∠ DEF$.
如图,当点$E$在$AD$左侧,且在$CD$下方时,
$\because AB// CD,\therefore ∠ A=∠ ADC$.
$\because EF// CD,\therefore ∠ DEF=∠ EDC$.
$\therefore ∠ A+∠ DEF=∠ ADE$.
如图,当点$E$在$AD$右侧,且在$CD$下方时,
$\because AB// CD,\therefore ∠ A=∠ ADC$.
$\because EF// CD,\therefore ∠ DEF=∠ EDC$.
$\therefore ∠ A+∠ DEF+∠ ADE=360°$.
综上所述,可得$∠ A-∠ DEF=∠ ADE$或$∠ A+∠ ADE=∠ DEF$或$∠ A+∠ DEF=∠ ADE$或$∠ A+∠ DEF+∠ ADE=360°$.