?AB?CD?∴在△ABF和△DCE中，?BF?CE，∴△ABF≌△DCE(SSS) ······························· 3分
?AF?DE?∴DE=BF ······································································································ 4分 (2)由（1）得△ABF≌△DCE ∴∠B=∠C ··························································· 5分 ∵平行四边形ABCD ∴AB∥CD ∴∠B+∠C=180° ········································· 6分 ∴∠B=∠C=90° ∴四边形ABCD是矩形. ° ······················································ 8分 22. （本题满分8分）
解：（1）50，86.4 （2）10（图略） （3）240 ·············································· 8分 23. （本题满分6分） 解：（1）16 ································································································ 2分
(2)
m 1 2 3 4 1 （2，1） （3，1） （4，1） n （1，1）2 （1，2） （2，2） （3，2） （4，2） 3 （1，3） （2，3） （3，3） （4，3） 4 （1，2） （2，4） （3，4） （4，4） ……………………………4分
∴所有可能的结果有16种等可能的结果，其中符合题意的有9种 ··························· 5分 ∴P?9 ········································································································ 6分
1624. （本题满分8分）
7
（1） y （2）
D B EOx
A
画出坐标系得1分，作出线段DE得2分 本小题4分（画对AB的两条垂线的给1分） 直线DE：y=
33x+得1分 2225. （本题满分8分）
解析：（1）由6x?4(x?1)得x?2，答：经2小时，后队追到前队 ·························· 2分
1 （2）先求联络员从后队出发到前队所用时间：由12x?4(x?1)得x?，
21y 当0?x?时，y?4?8x ------------------3分 42再求联络员第一次返回后队时的时间：
112由6x?12(x?)?12?得x?， 8322312故当?x?时，y?16x?8 ------------------4分
23图象如右------------------6分 21x O 32
2(3) 联络员从后队出发到前队前，由12x?6x?4(x?1)?12x得x?，-----------------7分
71110联络员第一次返回时，由?12?12(x?)?6x?16x?8得x? ------------------8分
2217210答：联络员从出发到他折返后第一次与后队相遇的过程中，当x为h或h时，他离前队的路程
717与他离后队的路程相等． 26. （本题满分10分）
（1）∵A（12，0），B（0，16）,∠AOB=90°∴AB?OA2?OB2?122?162?20 ∵E是AB的中点 ∴BE=10 ·············································································· 2分 ∵CE⊥AB ∴cos?ABO?∴
OBBE? ABBC161025? ∴t? ····················································································· 4分 202t4BCBE32?,∴BE= ························································· 6分 ABBO58
（2）有题意得DE⊥AO，,当t=4时，BC=8 由△BCE∽△BAO 得
∵DE∥BO ∴△ADE∽△AOB 得∴OD=12-
AEAD204? ∴AD= ··································· 7分 ABAO252049696= ∴D(，0) ································································· 8分 25252550200?t?（3）且t?8且t?12.5 ······························································· 10分 71327. （本题满分10分） 解：（1）45°； ···························································································· 2分 （2）如图，作PD⊥y轴，垂足为D，设l与x轴交于点E，
?1?m?1?m，设点P坐标为：（，n） ············· 3分 22∵PA=PC，∴PA2?PC2，即AE2?PE2?CD2?PD2，
由题意得，抛物线的对称轴为：x=
?1?m1?m?1?m?∴（ +1）2+n2=（n+m）2+?，解得：n= ?22?2??1?m1?m∴P 点的坐标为：（ ，）； ························································· 4分
22?1?m1?m（3）存在点Q ，∵P 点的坐标为：（ ，），
22PA2+PC2=AE2+PE2+CD2+PD2=
222∵AC2=1+m2，∴PA2+PC2=AC2，
∴∠APC=90°，∴△PAC 是等腰直角三角形， ∵以Q、B、C 为顶点的三角形与△PAC 相似， ∴△QBC 是等腰直角三角形，·········································································· 6分 ∴由题意可得满足条件的点Q 的坐标为：（-m，0）， ·········································· 7分