摘要:29.问题解决 解:方法一:如图(1-1).连接. 由题设.得四边形和四边形关于直线对称. ∴垂直平分.∴··········································· 1分 ∵四边形是正方形.∴ ∵设则 在中.. ∴解得.即················································ 3分 在和在中. . . ······································································· 5分 设则∴ 解得即················································································· 6分 ∴··································································································· 7分 方法二:同方法一.········································································· 3分 如图(1-2).过点做交于点.连接 ∵∴四边形是平行四边形. ∴ 同理.四边形也是平行四边形.∴ ∵ 在与中 ∴····························· 5分 ∵······························································ 6分 ∴································································································· 7分 类比归纳 (或),, ·········································································· 10分 联系拓广 ···································································································· 12分

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