1. 作图并写出结论：
如图,点$P$是$∠ AOB$的边$OA$上一点,请过点$P$画出$OA$,$OB$的垂线,分别交$BO$的延长线于点$M,N$,则线段的长表示点$P$到直线$BO$的距离;线段的长表示点$M$到直线$AO$的距离;线段$ON$的长表示点$O$到直线的距离;点$P$到直线$OA$的距离为.

1. 作图如所示，PN PM PN 0

2. 一个角的余角比它的补角的一半小 $20°$,求这个角的度数.

2. 设这个角的度数是$x$,则它的余角为$90°-x$,补角为$180°-x$.
$\because$这个角的余角比它的补角的一半小$20°$,
$\therefore 90°-x=\frac{1}{2}(180°-x)-20°$,解得$x=40°$.
故这个角的度数是$40°$.

3. (2024·南京玄武区期末)如图,直线 AB 和直线 CD交于点 E,EF,EG 分别平分$∠ AEC$和$∠ BEC$.
(1)求证:EF 与 EG 有什么位置关系,试说明原因;
(2)若$∠ AEF=66°$,求$∠ BEG$的度数.

3. (1)$\because ∠ AEC + ∠ BEC = 180°$, 且 EF, EG 分别平分$∠ AEC$和$∠ BEC$,
$\therefore ∠ FEC + ∠ CEG = \frac{1}{2}(∠ AEC + ∠ BEC) = \frac{1}{2} × 180° = 90°$,$\therefore EF ⊥ EG$.
(2)由(1)知,$EF ⊥ EG$,$\therefore ∠ FEG = 90°$.
$\because ∠ AEF = 66°$,$\therefore ∠ BEG = 180° - ∠ AEF - ∠ FEG = 180° - 90° - 66° = 24°$.

4. 数形结合思想 直线 A B, C D 相交于点 $O,OF ⊥$ CD 于点 O, 作射线 O E, 且 O C 在 $∠ AOE$ 的内部.(本题所研究的角均小于 $180°$)
(1) ①当 O E, O F 在如图(1)所示位置时, 若$∠ BOD=20°,∠ BOE=130°$, 求$∠ EOF$的度数；
②当 O E, O F 在如图(2)所示位置时, 若 O F平分$∠ BOE$, 试说明 O C 平分$∠ AOE$.
(2) 若$∠ AOF=2∠ COE$, 请直接写出$∠ BOE$与$∠ AOC$之间的数量关系.

4. (1)①$\because OF ⊥ CD$于点$O$,$\therefore ∠ DOF = 90°$.
$\because ∠ BOD = 20°$,$\therefore ∠ BOF = 90° - 20° = 70°$.
$\because ∠ BOE = 130°$,
$\therefore ∠ EOF = ∠ BOE - ∠ BOF = 130° - 70° = 60°$.
则$∠ EOF$的度数为$60°$.
②$\because OF$平分$∠ BOE$,
$\therefore ∠ EOF = ∠ BOF = \frac{1}{2}∠ EOB$.
$\because OF ⊥ CD$,$\therefore ∠ COF = 90°$,
$\therefore ∠ COE + ∠ EOF = ∠ AOC + ∠ BOF = 90°$,
$\therefore ∠ COE = ∠ AOC$,即$OC$平分$∠ AOE$.
(2)$3∠ AOC + 2∠ BOE = 270°$或$∠ AOC + 2∠ BOE = 270°$.
理由如下：
当点$E,F$在直线$AB$的同侧时,如图(1),
记$∠ COE = α$,则$∠ AOF = 2∠ COE = 2α$.
$\because OF ⊥ CD$,$\therefore ∠ COF = 90°$,
$\therefore ∠ EOF = 90° - α$,$∠ AOC = ∠ AOF - ∠ COF = 2α - 90°$①,
$\therefore ∠ BOE = 180° - ∠ AOC - ∠ COE = 180° - (2α - 90°) - α = 270° - 3α$②,
①$× 3 +$②$× 2$,得$3∠ AOC + 2∠ BOE = 270°$;
当点$E,F$在直线$AB$的异侧时,如图(2),
记$∠ COE = α$,则$∠ AOF = 2∠ COE = 2α$.
$\because OF ⊥ CD$,$\therefore ∠ COF = 90°$,
$\therefore ∠ AOC = ∠ COF - ∠ AOF = 90° - 2α$①,
$\therefore ∠ BOE = 180° - ∠ AOC - ∠ COE = 180° - (90° - 2α) - α = 90° + α$②,
$\therefore ∠ AOC + 2∠ BOE = 270°$.
综上可知,$3∠ AOC + 2∠ BOE = 270°$或$∠ AOC + 2∠ BOE = 270°$.