22. 如图，$a,b,c$ 三根木棒钉在一起，$∠ 1 = 70°$，$∠ 2 = 100°$，现将木棒 $a,b$ 同时分别以 $17°/\mathrm{s}$ 和 $2°/\mathrm{s}$ 的速度沿顺时针方向旋转，当木棒 $a$ 旋转一周时，两根木棒同时停止旋转，则旋转 $\_\_\_\_\_\_\mathrm{s}$ 后木棒 $a,b$ 平行.

22. 2或14 【点拨】本题考查平行线的判定.
【解析】设 $t$ 秒后木棒 $a ,b$ 平行,$100° - 17°t = 70° - 2°t$ ,解得 $t = 2$ ;
$180° + 100° - 17°t = 70° - 2°t$ ,解得 $t = 14$ . 综上所述,2 秒或 14 秒后木棒 $a ,b$ 平行. 故答案为 2 或 14.

23. 如图,将长方形纸片ABCD沿EF折叠,折痕EF交AD于点E,交BC于点F,点C,D的落点分别是点$C',D'$,$ED'$交BC于点G,再将四边形$C'D'GF$沿FG折叠,点$C',D'$的落点分别是点$C'',D''$,$GD''$交EF于点H. 下列四个结论:①$∠ GEF = ∠ GFE$,②$2∠ EFC = ∠ EGC + 180°$;③$∠ EGD'' = 2∠ EFG$;④$∠ EHG = 3∠ EFB$. 其中正确的结论是________.(填写序号)

23. ①②④ 【点拨】本题考查平行线的性质,解题关键是熟练掌握平行线的性质.
【解析】$\because AD // BC ,\therefore ∠DEF = ∠GFE$ . 根据折叠的性质得 $∠GEF = ∠DEF ,\therefore ∠GEF = ∠GFE$ . 故①正确,符合题意; $\because AD // BC ,\therefore ∠EGC + ∠GED = 180°$ . 根据折叠的性质得 $∠GED = 2∠DEF = 2∠GEF$ . $\because ∠EFC = 180° - ∠EFG = ∠EGC + ∠GEF ,\therefore 2∠EFC = 2∠EGC + 2∠GEF = ∠EGC + ∠GED + ∠EGC = ∠EGC + 180°$,故②正确,符合题意; $\because ∠D'GF = ∠GFE ,∠D'GF = 180° - ∠EGF = ∠GEF + ∠GFE ,\therefore ∠D'GF = 2∠EFG$ . 根据折叠的性质得 $∠D''GF = ∠D'GF$ . $\because ∠EGD'' + ∠D''GF + ∠D'GF = 180°,\therefore$ 当 $∠EGD'' = ∠D''GF = ∠D'GF = 60°$时,$∠EGD'' = 2∠EFG$ ,故③错误,不符合题意; $\because ∠D''GF = ∠D'GF = ∠EGB = ∠GED = ∠GEF + ∠FED = 2∠EFB ,\therefore ∠EHG = 180° - ∠GHF = ∠EFB + ∠D''GF = ∠EFB + 2∠EFB = 3∠EFB$ ,故④正确,符合题意. 故答案为①②④.

24. 如图,已知$AB// CD$,P为直线AB,CD外一点,BF平分$∠ABP$,DE平分$∠CDP$,BF的反向延长线交DE于点E.若$∠FED=α$,则用α表示$∠P$为________.

24. $∠ P=360°-2α$ 【点拨】本题考查平行线的性质,角平分线的定义,三角形内角和定理.
【解析】如图,延长 AB 交 PD 于点 G ,延长 FE 交 CD 于点 H . $\because BF$ 平分 $∠ ABP$ ,DE 平分 $∠ CDP ,\therefore ∠ 1 = ∠ 2 ,∠ 3 = ∠ 4 . \because AB // CD ,\therefore ∠ 1 = ∠ 5 ,∠ 6 = ∠ PDC = 2∠ 3 . \because ∠ PBG = 180° - 2∠ 1 ,\therefore ∠ PBG = 180° - 2∠ 5 ,\therefore ∠ 5 = 90° - \frac{1}{2} ∠ PBG . \because ∠ FED = 180° - ∠ HED ,∠ 5 = 180° - ∠ EHD ,∠ EHD + ∠ HED + ∠ 3 = 180° ,\therefore 180° - ∠ 5 + 180° - ∠ FED + ∠ 3 = 180° ,\therefore ∠ FED = 180° - ∠ 5 + ∠ 3 ,\therefore ∠ FED = 180° - ( 90° - \frac{1}{2} ∠ PBG ) + \frac{1}{2} ∠ 6 = 90° + \frac{1}{2} ( ∠ PBG + ∠ 6 ) = 90° + \frac{1}{2} ( 180° - ∠ P ) = 180° - \frac{1}{2} ∠ P . \because ∠ FED = α ,\therefore α = 180° - \frac{1}{2} ∠ P ,\therefore ∠ P = 360° - 2α$ . 故答案为 $∠ P = 360° - 2α$.

25. 如图,在平面直角坐标系中,点$A(-1,0)$,$B(2,3)$,$C(3,0)$,$D(a,-2a+1)$为第四象限内一点,连接$BD$交$x$轴于点$E$.若$S_{△ ADE}=S_{△ BCE}$,则$D$点坐标为$\underline{\hspace{5em}}$.

25. $( \frac{4}{3}, -\frac{5}{3} )$ 【点拨】本题考查平面直角坐标系中坐标与图形,解题的关键是将图形割补成规则的图形.
【解析】如题图,过点 B 向左作 $BH // x$ 轴,过点 A 作 $AH ⊥ BH$ ,过点 D 作 $DF ⊥ AH$ 交 HA 的延长线于点 F . 由题意得 $BH = AH = 3 ,AC = 4 ,AF = 2a - 1 ,DF = a + 1 . \because S_{△ ADE} = S_{△ BCE} ,\therefore S_{△ ADE} + S_{△ ABE} = S_{△ BCE} + S_{△ ABE} ,\therefore S_{△ ABD} = S_{△ ABC} = \frac{1}{2} · y_B · AC = \frac{1}{2} × 3 × 4 = 6 . \because S_{△ ABD} = S_{\mathrm{梯形}BDFH} - S_{△ AHB} - S_{△ AFD} = \frac{1}{2} (a + 1 + 3 )(3 + 2a - 1 ) - \frac{1}{2} × 3 × 3 - \frac{1}{2} × (2a - 1 )(a + 1 ),\therefore$ 解得 $a = \frac{4}{3} ,\therefore D$ 点坐标为 $( \frac{4}{3}, -\frac{5}{3} )$. 故答案为 $( \frac{4}{3}, -\frac{5}{3} )$.

五、解答题(本大题共3小题,共34分.解答应写出过程)
26. (10分)在平面直角坐标系$xOy$中,对于任意三点$A,B,C$的“矩面积”,给出如下定义:“水平底”$a$:任意两点横坐标差的最大值,“铅垂高”$h$:任意两点纵坐标差的最大值,则“矩面积”$S=ah$.例如:三点坐标分别为$A(1,2),B(-3,1),C(2,-2)$,则“水平底”$a=5$,“铅垂高”$h=4$,“矩面积”$S=ah=5×4=20$.根据所给定义,解决下列问题:
(1)若点$D(2,1),E(0,-3),F(4,1)$,则这三点“水平底”$a$的值为________;
(2)若点$D(\sqrt{5},2),E(-2\sqrt{5},\sqrt{2}),F(0,6+\sqrt{2})$,求这三点的“矩面积”;
(3)若$D(1,2),E(-2,1),F(0,t)$三点的“矩面积”为9,求点$F$的坐标.

26. 【点拨】本题考查新定义,解题关键是明确题目中的新定义.
【解析】(1)$\because D(2,1 ),E(0,-3 ),F(4,1 ),\therefore a = 4 - 0 = 4$ . 故答案为 4.
(2)$\because D(\sqrt{5},2 ),E(-2\sqrt{5},\sqrt{2} ),F(0,6+\sqrt{2} ),\therefore a = \sqrt{5} - (-2\sqrt{5} ) = 3\sqrt{5} ,h = 6 + \sqrt{2} - \sqrt{2} = 6 ,\therefore S = ah = 3\sqrt{5} ×6 = 18\sqrt{5}$.
(3)对于 $D(1,2 ),E(-2,1 ),F(0,t )$,其“水平底”$a = 3$.
$\because$ “矩面积”$S = 9 = ah ,\therefore h = 3$.
若 $t$ 为最大值,则 $h = t - 1 = 3 ,\therefore t = 4$ ;
若 $t$ 为最小值,则 $h = 2 - t = 3 ,\therefore t = -1$ ;
若 $1 ＜ t ＜ 2$ ,则 $h = 2 - 1 = 1 ≠ 3$.
综上所述,点 F 的坐标为$(0,4 )$或$(0,-1 )$.