27. (12 分)已知直线$AB// CD$，点$M,N$分别在直线$AB,CD$上，$H$为平面内一点，连接$HM,HN$.
(1)如图1，延长$HN$至点$G$，$∠ BMH$和$∠ GND$的平分线相交于点$E$.
①若$∠ BME=25°$，$∠ END=75°$，则$∠ H$的度数为________；
②探究$∠ MEN$与$∠ H$的数量关系，并给予证明；
(2)如图2，$∠ BMH$与$∠ HND$的平分线相交于点$E$，作$MP$平分$∠ AMH$交$CD$于点$P$，$NQ// MP$交$ME$的延长线于点$Q$.若$∠ H=150°$，求$∠ ENQ$的度数.

27. 【点拨】本题考查平行线的性质、角平分线的性质、邻补角、等量代换、角度之间的数量关系运算、辅助线的作法,解题关键是正确作出辅助线.
【解析】(1)①$\because ∠BMH$ 和 $∠GND$ 的平分线相交于点 E ,
$\therefore ∠BMH = 2∠BME = 50°,∠GND = 2∠EDN = 150°$,
$\therefore ∠DNH = 180° - 150° = 30°. \because AB // CD ,\therefore ∠DOH = ∠BMH = 50°$,
$\therefore ∠H = 180° - ∠DNH - (180° - ∠DOH ) = ∠DOH - ∠DNH = 50° - 30° = 20°$. 故答案为 20°.
②$2∠MEN - ∠H = 180°$.
证明:如图1,过点 E 作 $EP // AB$ 交 MH 于点 Q .
$\because EP // AB ,AB // CD ,\therefore EP // CD$.
又$\because ∠BMH$ 和 $∠GND$ 的平分线相交于点 E ,$\therefore ∠QEN = ∠DNE = \frac{1}{2} ∠GND ,∠MEQ = ∠BME = \frac{1}{2} ∠BMH$ ,
$\therefore ∠MEN = ∠MEQ + ∠QEN = \frac{1}{2} ∠BMH + \frac{1}{2} ∠GND = \frac{1}{2} ( ∠BMH + ∠GND ),\therefore 2∠MEN = ∠BMH + ∠GND$.
$\because ∠GND + ∠DNH = 180°,∠DNH + ∠H = 180° - ∠NOH = ∠MON = ∠BMH$ ,
$\therefore ∠GND + ∠BMH - ∠H = 180°,\therefore 2∠MEN - ∠H = 180°$.

(2)如图2,过点 H 作 $GI // AB$ ,交 MP 于点 G ,过点 H 作 $HT // MP$ ,交 AB 于点 T . 由(1)可得 $∠MEN = \frac{1}{2} ( ∠BMH + ∠HND )$.
由图可知 $∠MHN = ∠MHI + ∠NHI$.
$\because GI // AB ,\therefore ∠AMH = ∠MHI = 180° - ∠BMH$.
$\because GI // AB ,AB // CD ,\therefore GI // CD$ ,
$\therefore ∠HNC = ∠NHI = 180° - ∠HND$.
$\therefore ∠AMH + ∠HNC = 180° - ∠BMH + 180° - ∠HND = 360° - ( ∠BMH + ∠HND )$.
又$\because ∠AMH + ∠HNC = ∠MHI + ∠NHI = ∠MHN$ ,
$\therefore ∠BMH + ∠HND = 360° - ∠MHN$ ,即 $2∠MEN + ∠MHN = 360°$.
$\because ∠MHN = 150°,∴ ∠MEN = ( 360° - 150° ) ÷ 2 = 105°$.
$\because MP // NQ ,\therefore HT // NQ ,\therefore ∠ENQ + ∠ENH + ∠NHT = 180°$.
$\because MP$ 平分 $∠AMH ,\therefore ∠PMH = \frac{1}{2} ∠AMH = \frac{1}{2} ( 180° - ∠BMH )$.
$\because ∠NHT = ∠MHN - ∠MHT = 150° - ∠PMH$ ,
$\therefore ∠ENQ + ∠ENH + 150° - \frac{1}{2} ( 180° - ∠BMH ) = 180°$.
$\because ∠ENH = \frac{1}{2} ∠HND ,\therefore ∠ENQ + \frac{1}{2} ∠HND + 150° - 90° + \frac{1}{2} ∠BMH = 180°$,
$\therefore ∠ENQ + \frac{1}{2} ( ∠HND + ∠BMH ) = 120°$,
$\therefore ∠ENQ + ∠MEN = 120°,∴ ∠ENQ = 120° - 105° = 15°$.