如圖,平行四邊形ABCD中,AB=5,BC=10,BC邊上的高AM=4,E為 BC邊上的一個動點(不與B、C重合).過E作直線AB的垂線,垂足為F. FE與DC的延長線相交于點G,連結DE,DF..
1.求證:ΔBEF∽ΔCEG.
2.當點E在線段BC上運動時,△BEF和△CEG的周長之間有什么關系?并說明你的理由.
3.設BE=x,△DEF的面積為 y,請你求出y和x之間的函數關系式,并求出當x為何值時,y有最大值,最大值是多少?
1.因為四邊形ABCD是平行四邊形, 所以 ···················································· 1分
所以
所以
2.的周長之和為定值.··································································· 4分
理由一:
過點C作FG的平行線交直線AB于H ,
因為GF⊥AB,所以四邊形FHCG為矩形.所以 FH=CG,FG=CH
因此,的周長之和等于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與Rt△GCE中,有:
,
所以,△BEF的周長是, △ECG的周長是
又BE+CE=10,因此的周長之和是24.
3.設BE=x,則
所以 ···································· 8分
配方得:.
所以,當時,y有最大值.············································································ 10分
最大值為.
解析:略
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