?xM?xF?0，xN?xE?0，
?点M与点F对称，点N与点E对称． ····································································· 8分
1y （3）?a??0．
2P ?抛物线y1开口向下，抛物线y2开口向上． ·················· 9分
根据题意，得CD?y1?y2
A C O Q D x B 11?1??1?·················································· 11分 ???x2?x?1???x2?x?1???x2?2． ·
22?2??2?·················································· 12分 ?xA≤x≤xB，?当x?0时，CD有最大值2． ·
说明：第（2）问中，结论写成“M，N，或“MN?EF”E，F四点横坐标的代数和为0”
均得1分．
26.（08江西南昌）25．如图1，正方形ABCD和正三角形EFG的边长都为1，点E，F分别在线段AB，AD上滑动，设点G到CD的距离为x，到BC的距离为y，记?HEF为?（当点E，F分别与B，A重合时，记?（1）当?． ?0?）
，求x； ?0?时（如图2所示），y的值（结果保留根号）
（2）当?为何值时，点G落在对角形AC上？请说出你的理由，并求出此时x，y的值（结果保留根号）；
（3）请你补充完成下表（精确到0.01）：
? x y 0? 15? 0.03 0.29 30? 0 0.13 45? 60? 75? 0.29 0.03 90? （4）若将“点E，F分别在线段AB，AD上滑动”改为“点E，F分别在正方形ABCD边上滑动”．当滑动一周时，请使用（3）的结果，在图4中描出部分点后，勾画出点G运动所形成的大致图形．
31
（参考数据：
H A
E B F 3≈1.732，sin15??6?26?2） ≈0.259，sin75??≈0.966．
44H A
D
H D A(F) G C B(E) 图2
H D A G C B D
图1
C 图3
B
图4
C
（08江西南昌25题解析）25．解：（1）过G作MN?AB于M交CD于N，GK于K．
?BC??ABG?60?，BG?1，
?MG?13，BM?． ····························································································· 2分
2213，y?． ································································································· 3分
22············································ 4分 ?45?时，点G在对角线AC上，其理由是： ·
H A(F) M B(E) D G N K C ?x?1?（2）当?过G作IQ∥BC交AB，CD于I，Q， 过G作JP∥AB交AD，BC于J，P．
?AC平分?BCD，?GP?GQ，?GI?GJ．
?GE?GF，?Rt△GEI≌Rt△GFJ，??GEI??GFJ．
??GEF??GFE?60?，??AEF??AFE． ??EAF?90?，??AEF??AFE?45?．
即?····································································· 6分 ?45?时，点G落在对角线AC上． ·
H A E I B F J D
（以下给出两种求x，y的解法） 方法一：??AEG?45??60??105?，??GEI?75?．
?6?2在Rt△GEI中，GI?GE?， sin75?4G Q P C 32
?GQ?IQ?GI?1?6?2． ················································································· 7分 4?x?y?1?6?2． ································································································ 8分 4方法二：当点G在对角线AC上时，有
13·································································································· 7分 ??2x?2， ·
22解得x?1?6?2 46?2． ································································································ 8分 415?
0.03 0.
0.
0 0.
?x?y?1?（3）
?
x
0
0.13
?30?
45? 0.03
0.
13
60? 0.
29 0
75? 0.
50 0.
90?0.0.
50 29 13 03 03 13
······································································ 10分 （4）由点G所得到的大致图形如图所示： H A
D
y
B
C ··································································································· 12分
说明：1．第（2）问回答正确的得1分，证明正确的得2分，求出x ，y的值各得1分；2．第（3）问表格数据，每填对其中4空得1分；
3．第（4）问图形画得大致正确的得2分，只画出图形一部分的得1分． 27.（08山东滨州）23、（1）探究新知：如图1，已知△ABC与△ABD的面积相等，试判断AB与CD的位置关系，并说明理由.
33
CDAB
（2）结论应用：①如图2，点M、N在反比例函数y=
k(k?0)的图象上，过点M作MEx⊥y轴，过点N作NF⊥x轴，垂足分别为E，F. 试应用（1）中得到的结论证明：MN∥EF.
yEMNOFx
②若①中的其他条件不变，只改变点M，N的位置如图3所示，请判断MN与E是否
平行.
yMOxN （08山东滨州23题解析）23．（1）证明：分别过点C、D作CG?AB、DH垂足为G、H，则?CGA??DHB?90.
0?AB.
34
?CG?DH??ABC与?ABD的面积相等 ?CG=DH?四边形CGHD为平行四边形?AB?CD.（2）①证明：连结MF，NE
设点M的坐标为(x1,y1)，点N的坐标为(x2,y2)， ∵点M，N在反比例函数y?∴x1y1?k，x2y2k?k?0?的图象上， x?k
?ME?y轴，NF?x轴
?OE=y1,OF?x211?S?EFM?x1y1?k2211S?EFN?x2y2?k
22?S?EFM?S?EFN由（1）中的结论可知：MN∥EF。 ②MN∥EF。
28.（08山东滨州）24．（本题满分12分）
如图（1），已知在?ABC中，AB=AC=10，AD为底边BC上的高，且AD=6。将?ACD沿箭头所示的方向平移，得到?ACD。如图（2），AD交AB于E，AC分别交AB、AD于G、F。以D//////D为直径作?O，设BD/的长为x，?O的面积为y。
（1）求y与x之间的函数关系式及自变量x的取值范围； （2）连结EF，求EF与?O相切时x的值；
35