执笔抒情············································································10分 BE2 CF2 AD2 BD2 CE2 AF2 ·
BE2 CF2 AD2 a AD a BE a CF
········································11分 a2 2ADa AD2 a2 2BEa BE2 a2 2CFa CF2 ·整理得：2a AD BE CF 3a
2
222
AD BE CF
3
·····································································································12分 a ·2
1 4
25．（1）解:方法一，如图1，当x 1时,y 当x 4时,y 4
∴A 1············································································· 1分 ·
1 4
（图1）
B 4，4 ······················································································ 2分
设直线AB的解析式为y kx b ·············································· 3分
13
k b k 则 4 解得 4 4k b 4 b 1
∴直线AB的解析式为y 当x 0时,y 1
3
x 1 ············································· 4分 4
F 01，···························································································································5分 ·
方法二:求A、B两点坐标同方法一,如图2,作FG BD,AH BD,垂足分别为G、H,交y轴于点N,则四边形FOMG和四边形
NOMH均为矩形,设FO x···················································· 3分 △BGF∽△BHA
BGFG （图2） BHAH4 x4
·······················································································································4分 ·
54 4解得x 1
F 0，1 ·························································································································5分
