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23. (10分)已知$AB// CD$，E，F分别在AB，CD上.
(1)点G在AB，CD之间，连接EG，FG.
①如图1，若$∠ BEG = ∠ CFG - 90°$，求$∠ EGF$的度数；
②如图2，分别过点E，F的射线交于直线CD下方的点H，若$∠ HEG = 2∠ AEH$，$∠ DFG = 3∠ CFH$，求$∠ EGF$与$∠ EHF$的数量关系；
(2)如图3，射线EM从EB开始，绕点E以$20°$每秒的速度顺时针旋转，同时射线FN从FC开始，绕点F以$50°$每秒的速度顺时针旋转，直线ME与直线NF交于点P，若直线ME与直线NF相交所夹的锐角为$30°$，直接写出运动时间$t$秒($0≤ t≤8$)的值.
(1)点G在AB，CD之间，连接EG，FG.
①如图1，若$∠ BEG = ∠ CFG - 90°$，求$∠ EGF$的度数；
②如图2，分别过点E，F的射线交于直线CD下方的点H，若$∠ HEG = 2∠ AEH$，$∠ DFG = 3∠ CFH$，求$∠ EGF$与$∠ EHF$的数量关系；
(2)如图3，射线EM从EB开始，绕点E以$20°$每秒的速度顺时针旋转，同时射线FN从FC开始，绕点F以$50°$每秒的速度顺时针旋转，直线ME与直线NF交于点P，若直线ME与直线NF相交所夹的锐角为$30°$，直接写出运动时间$t$秒($0≤ t≤8$)的值.
答案
23. 【点拨】本题考查平行线的判定与性质,解题的关键是辅助线构造平行,注意情况进行讨论.
【解析】(1)①如题图1,向左作$GH // AB$,
$\because AB // CD, \therefore GH // CD // AB$,
$\therefore ∠EGH + ∠FGH = ∠BEG + ∠CFG = 180°, \because ∠BEG = ∠CFG - 90°$,
$\therefore ∠EGH = ∠CFG - 90°, ∠FGH = 180° - ∠CFG, \therefore ∠EGF = ∠EGH + ∠FGH = ∠CFG - 90° + 180° - ∠CFG = 90°$;
②如题图2,向左作$GK // AB$,作$HJ // CD$.
$\because AB // CD, \therefore GK // CD // AB // HJ$, 设$∠AEH = x°, ∠CFH = y°$, 则$∠HEG = 2∠AEH = 2x°, ∠DFG = 3∠CFH = 3y°, \because AB // HJ, CD // HJ, \therefore ∠EHJ = ∠AEH = x°, ∠FHJ = ∠CFH = y°, \therefore ∠EHF = ∠EHJ - ∠FHJ = x° - y°, \therefore ∠EGK = 180° - ∠AEG = 180° - 3x°, ∠FGK = ∠DFG = 3y°, \therefore ∠EGF = ∠EGK + ∠FGK = 180° - 3x° + 3y° = 180° - 3(x° - y°), \therefore ∠EGF = 180° - 3∠EHF, \therefore ∠EGF + 3∠EHF = 180°.$
(2)分三种情况进行讨论:
①如图1,作$PT // AB$. $\because AB // CD, \therefore PT // CD // AB, \therefore ∠TPF = ∠PFC = 50°t, ∠TPE = ∠PEA = 20°t, \therefore 20°t + 30° = 50°t$, 解得 $t = 1$.
②如图2,作$PT // AB$. 则 $PT // CD // AB, \therefore ∠BEP = ∠TPM = 20°t, ∠TPN = ∠DFN = 50°t - 180°, \therefore 20°t - (50°t - 180°) = 30°$, 解得 $t = 5$.
③如图3,作$PT // AB$. $\because ∠MPT = ∠BEM = 20°t, ∠FPT = ∠CFN = 360° - 50°t, ∠NPM = 30°, \therefore 20°t + 360° - 50°t + 30° = 180°$, 解得 $t = 7$.
综上可知,$t = 1$ 或 $t = 5$ 或 $t = 7$.
24. (12分)如图1,在平面直角坐标系中,点$A(a,0)$,点$B(b,0)$,点$C(0,c)$,且$a,b,c$满足$(a+3)^2 = \sqrt{b-2} + \sqrt{2-b} - |c-4|$.
(1)点$A,B,C$坐标分别是$A$
(2)如图2,点$K(m,n)$为线段$BC$上一点,$0≤ m≤2$.若直线$KO// AC$,求点$K$的坐标;
(3)如图3,点$T(0,t)$为$y$轴上的一动点.在点$T$运动过程中,若$S_{△ ABT}≤ S_{△ CBT}$,求$t$的取值范围.
(1)点$A,B,C$坐标分别是$A$
$(-3,0)$
,$B$$(2,0)$
,$C$$(0,4)$
;(2)如图2,点$K(m,n)$为线段$BC$上一点,$0≤ m≤2$.若直线$KO// AC$,求点$K$的坐标;
(3)如图3,点$T(0,t)$为$y$轴上的一动点.在点$T$运动过程中,若$S_{△ ABT}≤ S_{△ CBT}$,求$t$的取值范围.
答案
24. 【点拨】本题考查非负性,平行线的性质,一次函数解析式及三角形的面积公式等,灵活运用所学知识求解即可.
【解析】(1)$\because a,b,c$ 满足 $(a+3)^2 = \sqrt{b-2} + \sqrt{2-b} - |c-4|, \therefore b-2 ≥ 0, 2-b ≥ 0, \therefore b = 2$. 当 $b = 2$ 时, $(a+3)^2 = -|c-4|, \therefore a+3 = 0, c-4 = 0, \therefore a = -3, c = 4, \therefore$ 点 $A$ 的坐标为 $(-3,0)$, 点 $B$ 的坐标为 $(2,0)$, 点 $C$ 坐标为 $(0,4)$.
故答案为 $(-3,0),(2,0),(0,4)$.
(2)$\because$ 点 $A(-3,0),B(2,0),C(0,4), \therefore AO = 3, BO = 2, CO = 4$.
如题图,连接 $AK, \because OK // AC, \therefore S_{△ AOK} = S_{△ COK}, \therefore S_{△ AOK} + S_{△ BOK} = S_{△ COK} + S_{△ BOK}$,
即 $S_{△ ABK} = S_{△ BOC}, \therefore \frac{1}{2} AB · y_k = \frac{1}{2} × 2 × 4, \therefore y_k = \frac{8}{5}. \because S_{△ AOK} = S_{△ COK}$,
$\therefore \frac{1}{2} AO · y_k = \frac{1}{2} CO · x_k$, 即 $\frac{1}{2} × 3 × \frac{8}{5} = \frac{1}{2} × 4 · x_k, \therefore x_k = \frac{6}{5}$,
$\therefore K$ 点的坐标为 $(\frac{6}{5},\frac{8}{5})$.
(3)如题图,连接 $AT,BT,A(-3,0),B(2,0)$,
$\therefore AB = 5, OB = 2, \therefore S_{△ ABT} = \frac{1}{2} AB · |t| = \frac{5}{2} |t|$.
$\because C(0,4),T(0,t), \therefore TC = |4 - t|, \therefore S_{△ CBT} = \frac{1}{2} OB · |4 - t| = \frac{1}{2} × 2 × |4 - t| = |4 - t|. \because S_{△ ABT} ≤ S_{△ CBT}, \therefore \frac{5}{2} |t| ≤ |4 - t|$.
当 $t ＜ 0$ 时, $-\frac{5}{2}t ≤ 4 - t$, 解得 $t ≥ -\frac{8}{3}, \therefore -\frac{8}{3} ≤ t ＜ 0$; 当 $0 ＜ t ≤ 4$ 时, $\frac{5}{2}t ≤ 4 - t$, 解得 $t ≤ \frac{8}{7}, \therefore 0 ＜ t ≤ \frac{8}{7}$; 当 $t ＞ 4$ 时, $\frac{5}{2}t ≤ t - 4$, 解得 $t ≤ -\frac{8}{3}, \therefore$ 无解. 综上所述,$t$ 的取值范围为 $-\frac{8}{3} ≤ t ＜ 0$ 或 $0 ＜ t ≤ \frac{8}{7}$.
【解析】(1)$\because a,b,c$ 满足 $(a+3)^2 = \sqrt{b-2} + \sqrt{2-b} - |c-4|, \therefore b-2 ≥ 0, 2-b ≥ 0, \therefore b = 2$. 当 $b = 2$ 时, $(a+3)^2 = -|c-4|, \therefore a+3 = 0, c-4 = 0, \therefore a = -3, c = 4, \therefore$ 点 $A$ 的坐标为 $(-3,0)$, 点 $B$ 的坐标为 $(2,0)$, 点 $C$ 坐标为 $(0,4)$.
故答案为 $(-3,0),(2,0),(0,4)$.
(2)$\because$ 点 $A(-3,0),B(2,0),C(0,4), \therefore AO = 3, BO = 2, CO = 4$.
如题图,连接 $AK, \because OK // AC, \therefore S_{△ AOK} = S_{△ COK}, \therefore S_{△ AOK} + S_{△ BOK} = S_{△ COK} + S_{△ BOK}$,
即 $S_{△ ABK} = S_{△ BOC}, \therefore \frac{1}{2} AB · y_k = \frac{1}{2} × 2 × 4, \therefore y_k = \frac{8}{5}. \because S_{△ AOK} = S_{△ COK}$,
$\therefore \frac{1}{2} AO · y_k = \frac{1}{2} CO · x_k$, 即 $\frac{1}{2} × 3 × \frac{8}{5} = \frac{1}{2} × 4 · x_k, \therefore x_k = \frac{6}{5}$,
$\therefore K$ 点的坐标为 $(\frac{6}{5},\frac{8}{5})$.
(3)如题图,连接 $AT,BT,A(-3,0),B(2,0)$,
$\therefore AB = 5, OB = 2, \therefore S_{△ ABT} = \frac{1}{2} AB · |t| = \frac{5}{2} |t|$.
$\because C(0,4),T(0,t), \therefore TC = |4 - t|, \therefore S_{△ CBT} = \frac{1}{2} OB · |4 - t| = \frac{1}{2} × 2 × |4 - t| = |4 - t|. \because S_{△ ABT} ≤ S_{△ CBT}, \therefore \frac{5}{2} |t| ≤ |4 - t|$.
当 $t ＜ 0$ 时, $-\frac{5}{2}t ≤ 4 - t$, 解得 $t ≥ -\frac{8}{3}, \therefore -\frac{8}{3} ≤ t ＜ 0$; 当 $0 ＜ t ≤ 4$ 时, $\frac{5}{2}t ≤ 4 - t$, 解得 $t ≤ \frac{8}{7}, \therefore 0 ＜ t ≤ \frac{8}{7}$; 当 $t ＞ 4$ 时, $\frac{5}{2}t ≤ t - 4$, 解得 $t ≤ -\frac{8}{3}, \therefore$ 无解. 综上所述,$t$ 的取值范围为 $-\frac{8}{3} ≤ t ＜ 0$ 或 $0 ＜ t ≤ \frac{8}{7}$.
