11. 如图,$∠AOB=α$,OC平分$∠AOB$,D是边OA上一点,将射线OB沿OD平移至射线DE,交OC于点F,点E在点F的右侧.M是射线DA上一点(与点D不重合),N是线段DF上一点(与点D,F不重合),连接MN,$∠OMN=β$.
(1)请在图中根据题意补全图形；
(2)求$∠MNE$的度数(用含$α,β$的式子表示)；
(3)点G在线段OF上(与点O,F不重合),连接GN并延长交OA于点T,且满足$2∠NGO+∠OMN=180°$,画出符合题意的图形,并探究$∠ENM$与$∠ENG$之间的数量关系.
答案：11.解:(1)补全图形如答图所示.
(2)$\because DE// OB$,$∠ AOB=α$,
$\therefore ∠ ADE=∠ AOB=α$.
$\because ∠ MNE+∠ MND=180°$,$∠ ADE+∠ OMN+$
$∠ MND=180°$,
$\therefore ∠ MNE=∠ ADE+∠ OMN=α +β$.
(3)画出图形如答图.$∠ ENM=180°-2∠ ENG$.
$\because OC$平分$∠ AOB$,
$\therefore ∠ AOC=∠ BOC=\frac{1}{2}α$.
设$∠ NGO=\gamma$.
$\because ∠ ENM=α +β$,$2∠ NGO+∠ OMN=180°$,
$\therefore ∠ ENM=180°+α -2\gamma$.
$\because ∠ AOC+∠ NGO+∠ OTN=180°$,$∠ TDN+$
$∠ DNT+∠ OTN=180°$,
$\therefore ∠ AOC+∠ NGO=∠ TDN+∠ DNT$,
即$\frac{1}{2}α +\gamma =α +∠ DNT$,$\therefore ∠ DNT=\gamma -\frac{1}{2}α$.
$\because ∠ ENG=∠ DNT$,$\therefore ∠ ENG=\gamma -\frac{1}{2}α$,
$\therefore ∠ ENM=180°-2∠ ENG$.
