发布时间 : 星期三 文章中考数学压轴题100题精选（1-10题）更新完毕开始阅读d3f61f52e43a580216fc700abb68a98270feacee
∴△BAD≌△CBE…………………………………………2分 ∴AD=BE……………………………………………………3分 （2）∵E是AB中点，
∴EB=EA由（1）AD=BE得：AE=AD……………………………5分 ∵AD∥BC∴∠7=∠ACB=45°∵∠6=45°∴∠6=∠7 由等腰三角形的性质，得：EM=MD，AM⊥DE。 即，AC是线段ED的垂直平分线。……………………7分 （3）△DBC是等腰三角（CD=BD）……………………8分 理由如下：
由（2）得：CD=CE由（1）得：CE=BD∴CD=BD ∴△DBC是等腰三角形。……………………………10分 【009】解：（1）①QAC⊥x轴，AE⊥y轴，
y N E D A ?四边形AEOC为矩形．
QBF⊥x轴，BD⊥y轴， ?四边形BDOF为矩形．
B K O C F M x QAC⊥x轴，BD⊥y轴，
?四边形AEDK，DOCK，CFBK均为矩形． ············· 1分
图1 QOC?x1，AC?y1，x1gy1?k， ?S矩形AEOC?OCgAC?x1gy1?k QOF?x2，FB?y2，x2gy2?k， ?S矩形BDOF?OFgFB?x2gy2?k． ?S矩形AEOC?S矩形BDOF．
QS矩形AEDK?S矩形AEOC?S矩形DOCK，
S矩形CFBK?S矩形BDOF?S矩形DOCK，
?S矩形AEDK?S矩形CFBK． ··········································································································· 2分
②由（1）知S矩形AEDK?S矩形CFBK．
?AKgDK?BKgCK． AKBK??． ·························································································································· 4分 CKDKQ?AKB??CKD?90°， ?△AKB∽△CKD． ·············································································································· 5分 ??CDK??ABK． ?AB∥CD． ···························································································································· 6分
QAC∥y轴，
?四边形ACDN是平行四边形． ?AN?CD． ···························································································································· 7分 同理BM?CD．
··························································································································· 8分 ?AN?BM． ·
（2）AN与BM仍然相等． ····································································································· 9分
QS矩形AEDK?S矩形AEOC?S矩形ODKC，
S矩形BKCF?S矩形BDOF?S矩形ODKC，
y 又QS矩形AEOC?S矩形BDOF?k，
E N F M A ?S矩形AEDK?S矩形BKCF． ····································· 10分
x O C ?AKgDK?BKgCK．
D K CKDK B?． ?AKBK图2 Q?K??K，
?△CDK∽△ABK． ??CDK??ABK． ?AB∥CD． ·························································································································· 11分
QAC∥y轴，
?四边形ANDC是平行四边形． ?AN?CD． 同理BM?CD． ?AN?BM． ························································································································· 12分
??3a?4a?2b?3，?【010】解：（1）根据题意，得?b ····· 2分
??1.??2ay D E ?a?1，?抛物线对应的函数表达式为y?x2?2x?3． 3分 解得??b??2.（2）存在．
在y?x?2x?3中，令x?0，得y??3． 令y?0，得x?2x?3?0，?x1??1，x2?3．
22N A O 1 N x F C P ?A(?1，0)，B(3，0)，C(0，?3)．
M （第26题图）
2
又y?(x?1)?4，?顶点M(1················································································· 5分 ，?4)． ·容易求得直线CM的表达式是y??x?3． 在y??x?3中，令y?0，得x??3．
········································································································ 6分 ?N(?3，0)，?AN?2． ·
2在y?x?2x?3中，令y??3，得x1?0，x2?2．
?CP?2，?AN?CP．
····································· 8分 ?3)． ·QAN∥CP，?四边形ANCP为平行四边形，此时P(2，（3）△AEF是等腰直角三角形．
理由：在y??x?3中，令x?0，得y?3，令y?0，得x?3．
?直线y??x?3与坐标轴的交点是D(0，3)，B(3，0)．
···························································································· 9分 ?OD?OB，??OBD?45°． ·
又Q点C(0，························································· 10分 ?3)，?OB?OC．??OBC?45°． ·由图知?AEF??ABF?45°，?AFE??ABE?45°． ··············································· 11分
····································· 12分 ??EAF?90°，且AE?AF．?△AEF是等腰直角三角形． ·（4）当点E是直线y??x?3上任意一点时，（3）中的结论成立． 14分