如图10.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．——青夏教育精英家教网—

如图10.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? (1) 因为四边形ABCD是平行四边形. 所以 1分所以所以 ···························································································· 3分 (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．·········································· 6分 (3)设BE＝x.则所以 ···································· 8分配方得:．所以.当时.y有最大值．············································································· 9分最大值为．··········································································································· 10分【查看更多】

如图，在等腰梯形ABCD中，AD∥BC．O是CD边的中点，以O为圆心，OC长为半径作圆，交BC边于点E．过E作EH⊥AB，垂足为H．已知⊙O与AB边相切，切点为F．
（1）求证：OE∥AB；
（2）求证：EH=

1

2

AB；
（3）若AD与⊙O也相切，如图二，已知BE（BC）=5，BH=3，求⊙O的半径．
（江苏苏州10年中考27题改编）