题目

如图，平行四边形ABCD中，AB＝5，BC＝10，BC边上的高AM=4，E为 BC边上的一个动点（不与B、C重合）．过E作直线AB的垂线，垂足为F． FE与DC的延长线相交于点G，连结DE，DF．.1.求证：ΔBEF∽ΔCEG．2.当点E在线段BC上运动时，△BEF和△CEG的周长之间有什么关系？并说明你的理由．3.设BE＝x，△DEF的面积为 y，请你求出y和x之间的函数关系式，并求出当x为何值时,y有最大值，最大值是多少？  答案： 1.因为四边形ABCD是平行四边形， 所以 ···················································· 1分  所以所以2.的周长之和为定值．··································································· 4分理由一：过点C作FG的平行线交直线AB于H ，因为GF⊥AB，所以四边形FHCG为矩形．所以 FH＝CG，FG＝CH因此，的周长之和等于BC＋CH＋BH  由  BC＝10，AB＝5，AM＝4，可得CH＝8，BH＝6，所以BC＋CH＋BH＝24 ······························································································ 6分理由二：由AB＝5，AM＝4，可知   在Rt△BEF与Rt△GCE中，有：，所以，△BEF的周长是， △ECG的周长是又BE＋CE＝10，因此的周长之和是24．3.设BE＝x，则所以 ···································· 8分配方得：． 所以，当时，y有最大值．············································································ 10分最大值为． 解析:略 

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