23.已知∠AOC=20°,以O为顶点,OA为一边顺次往外画两个锐角∠AOB和∠AOD,并且∠BOD=2∠AOB,射线OM平分∠BOD,射线ON平分∠BOC.设∠AOB=α(20°＜α＜45°).
(1)如图8-8中,若∠AON=10°,求α的值；
(2)如图8-8中,若OE是∠BOM内的一条射线,且∠BOE=$\frac{3}{4}α -5°$,试说明OE是否平分∠DON.
图8-8

23.解:(1)$\because ∠ AON=10°,∠ AOC=20°$,
$\therefore ∠ NOC=∠ AOC+∠ AON=30°$.
$\because$射线ON平分$∠ BOC$,
$\therefore ∠ NOC=∠ BON=30°$,
$\therefore α=∠ AOB=∠ BON+∠ AON=40°$.
(2)射线OE平分$∠ DON$.理由如下:
$\because ∠ AOB=α,∠ AOC=20°$,
$\therefore ∠ BOC=α+20°$.
$\because$射线ON平分$∠ BOC$,
$\therefore ∠ BON=\dfrac{1}{2}∠ BOC=\dfrac{1}{2}α+10°$.
$\because ∠ BOD=2∠ AOB=2α$,OM平分$∠ BOD$,
$\therefore ∠ DOM=∠ BOM=∠ AOB=α$.
$\because ∠ BOE=\dfrac{3}{4}α-5°$,
$\therefore ∠ MOE=∠ BOM-∠ BOE=α-(\dfrac{3}{4}α-5°)=\dfrac{1}{4}α+5°$.
$\because ∠ DOE=∠ MOE+∠ DOM=\dfrac{1}{4}α+5°+α=\dfrac{5}{4}α+5°$,
$∠ NOE=∠ BOE+∠ BON=\dfrac{3}{4}α-5°+\dfrac{1}{2}α+10°=\dfrac{5}{4}α+5°$,
$\therefore ∠ DOE=∠ NOE$,
$\therefore$射线OE平分$∠ DON$.