题目

（12分）如图，在Rt△ABC中，∠C=90°，AC=BC=4cm，点D为AC边上一点，且AD=3cm，动点E从点A出发，以1cm/s的速度沿线段AB向终点B运动，运动时间为x s．作∠DEF=45°，与边BC相交于点F．设BF长为ycm．【小题1】（1）当x=   ▲ s时，DE⊥AB；【小题2】（2）求在点E运动过程中，y与x之间的函数关系式及点F运动路线的长；【小题3】（3）当△BEF为等腰三角形时，求x的值． 答案：【小题1】（1）【小题2】（2）∵在△ABC中，∠C=90°，AC=BC=4．∴∠A＝∠B＝45°，AB=4，∴∠ADE＋∠AED＝135°；又∵∠DEF＝45°，∴∠BEF＋∠AED＝135°，∴∠ADE＝∠BEF；∴△ADE∽△BEF···················································································· 4分∴＝，∴＝，∴y＝－x2+x························································ 5分∴y＝－x2+x＝－( x－2)2+∴当x＝2时，y有最大值＝·································································· 6分∴点F运动路程为cm············································································ 7分 【小题3】（3）这里有三种情况：①如图，若EF＝BF，则∠B＝∠BEF；又∵△ADE∽△BEF，∴∠A＝∠ADE＝45°∴∠AED＝90°，∴AE＝DE＝，∵动点E的速度为1cm/s，∴此时x＝s；②如图，若EF＝BE，则∠B＝∠EFB；又∵△ADE∽△BEF，∴∠A＝∠AED＝45°∴∠ADE＝90°，∴AE＝3，∵动点E的速度为1cm/s∴此时x＝3s； ③如图，若BF＝BE，则∠FEB＝∠EFB；又∵△ADE∽△BEF，∴∠ADE＝∠AED∴AE＝AD＝3，∵动点E的速度为1cm/s∴此时x＝3s；综上所述，当△BEF为等腰三角形时，x的值为s或3s或3s．（注：求对一个结论得2分，求对两个结论得4分，求对三个结论得5分）解析:略

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