《2025年假日数学吉林出版集团股份有限公司七年级人教版》

17. 如图,已知直线$AB$,$CD相交于点O$,$OE⊥AB$.
(1)如图①,请直接写出图中3对相等的角；(平角除外)
(2)如图②,作$OF平分∠COE$,求证：$∠EOD = 2∠AOF$；
(3)如图③,在(2)的条件下,点$M在OE$上,点$N在OD$上,连接$MN$,作$NP平分∠MNO交OF于点P$,交$OE于点Q$,若$∠EOF = 3∠BOD$,$∠PND - \frac{1}{2}∠AOF = 135^{\circ}$,求$∠OQN$的度数.

答案：

(1) 解：$\angle AOC = \angle BOD$，$\angle AOD =\angle BOC$，$\angle AOE = \angle BOE$.
(2) 证明：设$\angle BOD = \alpha$，
$\therefore \angle AOC = \angle BOD = \alpha$.
$\because OE \perp AB$，
$\therefore \angle AOE = \angle BOE = 90^{\circ}$，
$\therefore \angle EOD = 90^{\circ} - \alpha$.
$\because \angle COE + \angle EOD = 180^{\circ}$，
$\therefore \angle COE = 90^{\circ} + \alpha$.
$\because OF$平分$\angle COE$，
$\therefore \angle COF = \frac{1}{2}\angle COE = 45^{\circ} + \frac{1}{2}\alpha$，
$\therefore \angle AOF = \angle COF - \angle AOC = 45^{\circ} - \frac{1}{2}\alpha$，
$\therefore 2\angle AOF = 90^{\circ} - \alpha$，
$\therefore \angle EOD = 2\angle AOF$.
(3) 解：$\because NP$平分$\angle MNO$，
$\therefore \angle PNO = \angle MNP = \frac{1}{2}\angle MNO$，
$\therefore \angle PND = 180^{\circ} - \angle PNO = 180^{\circ} - \frac{1}{2}\angle MNO$.
$\because \angle PND - \frac{1}{2}\angle AOF = 135^{\circ}$，
$\therefore 180^{\circ} - \frac{1}{2}\angle MNO - \frac{1}{2}\angle AOF = 135^{\circ}$，
$\therefore \angle MNO + \angle AOF = 90^{\circ}$.
$\because \angle EOF + \angle AOF = 90^{\circ}$，
$\therefore \angle MNO = \angle EOF$.
$\because OF$平分$\angle COE$，
$\therefore \angle COF = \angle EOF = 3\angle BOD$，
$\therefore \angle MNO = \angle FOC$，$\therefore MN // OF$.
如图，作$QK // MN$，$\therefore QK // OF$，

$\therefore \angle NQK = \angle MNQ$，$\angle OQK = \angle POQ$.
$\therefore \angle OQN = \angle MNQ + \angle POQ$.
$\because \angle EOF = 3\angle BOD$，$OF$平分$\angle COE$，
$\therefore \angle COF = \angle EOF = 3\angle BOD$.
$\because \angle AOC = \angle BOD$，
$\therefore \angle AOF = 2\angle BOD$.
$\because \angle AOF + \angle EOF = 90^{\circ}$，
$\therefore 5\angle BOD = 90^{\circ}$.
$\therefore \angle BOD = 18^{\circ}$.
$\therefore \angle EOF = 54^{\circ}$. $\therefore \angle MNO = 54^{\circ}$.
$\therefore \angle MNQ = \frac{1}{2}\angle MNO = 27^{\circ}$.
$\therefore \angle OQN = 54^{\circ} + 27^{\circ} = 81^{\circ}$.