20. 如图10-5,直线BC与MN相交于点O,AO⊥BC.
(1)直接写出图中与∠AOM互余的角;
(2)已知OE平分∠BON,且∠EON=20°,求∠AOM的度数.

图10-5
(1)直接写出图中与∠AOM互余的角;
(2)已知OE平分∠BON,且∠EON=20°,求∠AOM的度数.
图10-5
答案
解:(1)与$∠ AOM$互余的角是:$∠ COM,∠ BON.$
(2)$\because OE$平分$∠ BON,$
$\therefore ∠ BON=2∠ EON=40°,$
$\therefore ∠ COM=∠ BON=40°.$
$\because AO⊥ BC,$
$\therefore ∠ AOC=90°,$
$\therefore ∠ AOM=90°-∠ COM=90°-40°=50°.$
(2)$\because OE$平分$∠ BON,$
$\therefore ∠ BON=2∠ EON=40°,$
$\therefore ∠ COM=∠ BON=40°.$
$\because AO⊥ BC,$
$\therefore ∠ AOC=90°,$
$\therefore ∠ AOM=90°-∠ COM=90°-40°=50°.$
21. 如图10-6,直线AB与CD被直线EF所截,EF与AB,CD分别交于点P,O,且$AO⊥BO$,$∠1+∠2=90°$。
(1)试说明:$AB// CD$;
(2)若OB平分$∠DOE$,$∠3=4∠2$,求$∠OPB$的度数。

(1)试说明:$AB// CD$;
(2)若OB平分$∠DOE$,$∠3=4∠2$,求$∠OPB$的度数。
答案
解:(1)$\because AO⊥ BO,$
$\therefore ∠ AOB=90°,$
$\therefore ∠ AOC+∠ 2=90°.$
$\because ∠ 1+∠ 2=90°,$
$\therefore ∠ AOC=∠ 1,$
$\therefore AB// CD.$
(2)$\because OB$平分$∠ DOE,$
$\therefore ∠ DOE=2∠ 2.$
$\because ∠ 3=4∠ 2,∠ 3+∠ DOE=180°,$
$\therefore 4∠ 2+2∠ 2=180°,$
$\therefore ∠ 2=30°,$
$\therefore ∠ DOE=60°.$
$\because AB// CD,$
$\therefore ∠ DOE+∠ OPB=180°,$
$\therefore ∠ OPB=180°-60°=120°.$
$\therefore ∠ AOB=90°,$
$\therefore ∠ AOC+∠ 2=90°.$
$\because ∠ 1+∠ 2=90°,$
$\therefore ∠ AOC=∠ 1,$
$\therefore AB// CD.$
(2)$\because OB$平分$∠ DOE,$
$\therefore ∠ DOE=2∠ 2.$
$\because ∠ 3=4∠ 2,∠ 3+∠ DOE=180°,$
$\therefore 4∠ 2+2∠ 2=180°,$
$\therefore ∠ 2=30°,$
$\therefore ∠ DOE=60°.$
$\because AB// CD,$
$\therefore ∠ DOE+∠ OPB=180°,$
$\therefore ∠ OPB=180°-60°=120°.$
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