∴ AB＝BC．
∵ △BEF是等边三角形，
∴ BF＝BE，∠FBE＝∠FEB＝60°．
∵ ∠ABC＝60°， ∴ ∠ABC＝∠FBE，
∴ ∠ABC－∠ABE＝∠FBE－∠ABE，即∠EBC＝∠FBA． ∴ △EBC≌△FBC（SAS）． ∴ CE＝AF． ············································································································ 4分
（2）解：∵ 四边形ABCD是菱形，
∴ AD∥BC，∠D＝∠ABC＝60°． ∴ ∠C＝180°－∠D＝120°．
在△PDE中，∠D＋∠DPE＋∠PED＝180°， ∴ ∠DEP＝72°．
由（1）得，∠FEB＝60°，
∴ ∠BED＝∠DEP＋∠BEP＝72°＋60°＝132°． ∴ ∠CBE＝∠BED－∠C＝132°－120°＝12°． ····················································· 8分
20．（本题8分）
（1）90，80，80． ··············································································································· 6分 （2）不合理，因为若将每位营销员月销售量定为90台，则多数营销员可能完不成任务． ················································································································································· 8分 21．（本题8分）
1
解：（1）3 ． ···················································································································· 2分 （2）随机选两位同学打第一场比赛，可能出现的结果有12种，即（甲，乙）、（甲，
丙）、（甲，丁）、（乙，甲）、（乙，丙），（乙，丁）、（丙，甲）、（丙，乙）、（丙，丁）、（丁，甲）、（丁，乙），（丁，丙）、并且它们出现的可能性相等．恰好选中甲、乙两位同学（记为事件A）的结果有2种，即（甲，乙）、（乙，甲），
21
所以P（A）＝12＝6． ···································································· 8分
22．（本题7分）
略 ········································································································································ 7分
23．（本题8分）
解：（1）12． ······················································································································ 2分 （2）设线段AD所表示的y与x之间的函数表达式为y＝kx＋b．
1
因为y＝kx＋b的图像过点（0，4）与（2，0），
??b＝4，?k＝－8，1所以? 解方程组，得?
b＝4．k＋b＝0．??2?
所以线段AD所表示的y与x之间的函数表达式为y＝－8x＋4． ··················· 5分
1（3）根据题意，联络员出发h后与第一次追上一班，此时，联络员与二班相距3 km，
2
31112
折返后需要12+6＝6(h)，因为＋6＝，
23
2
所以，联络员出发 h后与第一次后队相遇． ···················································· 8分
3
24．（本题8分）
证明：（1）如图，连接BD，交AC于点F．
∵ ∠BAD＝90°， ∴ BD是直径． ∴ ∠BCD＝90°． ∴ ∠DEC＋∠CDE＝90°． ∵ ∠DEC＝∠BAC， ∴ ∠BAC＋∠CDE＝90°． ∵ ∠BAC＝∠BDC， ∴ ∠BDC＋∠CDE＝90°． ∴ ∠BDE＝90°，即 BD⊥DE． ∵ 点D在⊙O上，
∴ DE是⊙O的切线． ·················································································· 4分
（2）∵ DE∥AC，∠BDE＝90°，
∴ ∠BFC＝90°．
1
∴ CB＝AB＝12，AF＝CF＝2AC，
∵ ∠CDE＋∠BDC＝90°，∠BDC＋∠CBD＝90°． ∴ ∠CDE＝∠CBD．
∵ ∠DCE＝∠BCD＝90°， ∴ △BCD∽△DCE， BCCD
∴ CD＝CE， ∴ CD＝6．∴ BD＝65．
CFCD125
同理：△CFD∽△BCD，∴ BC＝BD， ∴ CF＝5．
245
∴ AC＝2AF＝5． ·························································································· 8分
25．（本题8分）
解：设货车、客车的速度分别为x km/h、y km/h， 由题意，得AP＝PQ＝x km，BM＝MN＝y km. 如图，过点M作ME⊥AB，垂足为E． 在Rt△BME中， B 北 ME
东 ∵ sinB＝BM
E Q 26° A
D F O E C B （第24题）
37° M F 45° N P A C
