由题设，得四边形ABNM和四边形FENM关于直线MN对称．
∴MN垂直平分BE．∴BM?EM，BN?EN． ···································· 1分 ∵四边形ABCD是正方形，∴?A??D??C?90°,AB?BC?CD?DA?2． ∵
CE1设BN?x，则NE?x， ?，?CE?DE?1．NC?2?x．CD2222 在Rt△CNE中，NE?CN?CE．
55，即BN?． ········································· 3分 44 在Rt△ABM和在Rt△DEM中，
AM2?AB2?BM2， DM2?DE2?EM2，
?AM2?AB2?DM2?DE2． ····························································· 5分
2222 设AM?y，则DM?2?y，∴y?2??2?y??1．
11 解得y?，即AM?． ····································································· 6分
44AM1 ∴ ····················································································· 7分 ?．BN55 方法二：同方法一，BN?． ································································ 3分
4 如图（1－2），过点N做NG∥CD，交AD于点G，连接BE．
∴x??2?x??1．解得x?222
F G M A D
E
B C N
图（1-2）
∵AD∥BC，∴四边形GDCN是平行四边形． ∴NG?CD?BC． 同理，四边形ABNG也是平行四边形．∴AG?BN?5． 4 ∵MN?BE， ??EBC??BNM?90°． QNG?BC， ??MNG??BNM?90°，??EBC??MNG． 在△BCE与△NGM中
??EBC??MNG，? ?BC?NG，∴△BCE≌△NGM，EC?MG． ························· ５分
??C??NGM?90°．?∵AM?AG?MG，AM=∴
类比归纳
51 ····················································· 6分 ?1?．44AM1 ··················································································· 7分 ?．BN52?n?1? ·249（或）；； ································································ 10分 251017n?1联系拓广
n2m2?2n?1 ······················································································ 12分 22nm?1
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