11. (20分)已知直线$AB// CD$,直线EF与AB,CD分别交于点G,H,$∠ EHD=α(0°＜α＜90°)$.小安将一个含$60°$角的直角三角板PMN按如图①放置,使点N,M分别在直线AB,CD上,且在点G,H的右侧,$∠ P=90°$,$∠ PMN=60°$.
(1)填空:$∠ PNB+∠ PMD$
=
$∠ P$.
(2)若$∠ MNG$的平分线NO交直线CD于点O,如图②.
①当$NO// EF,PM// EF$时,求$α$的度数;
②小安将三角板PMN保持$PM// EF$并向左平移,在平移的过程中求$∠ MON$的度数.
答案：11. (1)$=$
(2)解:①$\because NO// EF$,$PM// EF$,
$\therefore NO// PM$,$\therefore ∠ ONM=∠ PMN$.
$\because ∠ PMN=60°$,$\therefore ∠ ONM=∠ PMN=60°$.
$\because NO$平分$∠ MNG$,
$\therefore ∠ ANO=∠ ONM=60°$.
$\because AB// CD$,$\therefore ∠ NOM=∠ ANO=60°$.
$\because NO// EF$,$\therefore ∠ EHD=∠ NOM=60°$,$\therefore α =60°$.
②当点$N$在点$G$的右侧时,如答图①,
$\because PM// EF$,$∠ EHD=α$,
$\therefore ∠ PMD=α$,$\therefore ∠ NMD=60°+α$.
$\because AB// CD$,$\therefore ∠ ANM=∠ NMD=60°+α$.
$\because NO$平分$∠ ANM$,
$\therefore ∠ ANO=\frac{1}{2}∠ ANM=30°+\frac{1}{2}α$.
$\because AB// CD$,$\therefore ∠ MON=∠ ANO=30°+\frac{1}{2}α$.
当点$N$在点$G$的左侧时,如答图②.
$\because PM// EF$,$∠ EHD=α$,
$\therefore ∠ PMD=α$,$\therefore ∠ NMD=60°+α$.
$\because AB// CD$,
$\therefore ∠ BNM+∠ NMO=180°$,$∠ BNO=∠ MON$.
$\because NO$平分$∠ MNG$,
$\therefore ∠ BNO=\frac{1}{2}∠ BNM=\frac{1}{2}[180°-(60°+α)]=60°-\frac{1}{2}α$,$\therefore ∠ MON=60°-\frac{1}{2}α$.
综上所述,$∠ MON$的度数为$30°+\frac{1}{2}α$或$60°-\frac{1}{2}α$.
