得
x?14?y3?35，?x?75，y?95，所以点D的坐标为??79，?55??． ?综上所述，点D的坐标为(3，3)或??79，55??······························································· （5分） ?． ·?25．解：（1）取AB中点H，联结MH，
?M为DE的中点，?MH∥BE，MH?12·································· （1分） (BE?AD)． ·
又?AB?BE，?MH?AB． ·················································································· （1分）
11·········································· （2分）（1分） ?S△ABM?AB?MH，得y?x?2(x?0)；
22（2）由已知得DE?······································································· （1分） (x?4)?2． ·
22?以线段AB为直径的圆与以线段DE为直径的圆外切，
?MH?1243AB?12DE，即
12(x?4)?431?2??222··························· （2分） (4?x)?2?． ·?解得x?，即线段BE的长为； ············································································· （1分）
（3）由已知，以A，N，D为顶点的三角形与△BME相似，
又易证得?DAM??EBM． ······················································································ （1分） 由此可知，另一对对应角相等有两种情况：①?ADN??BEM；②?ADB??BME． ①当?ADN??BEM时，?AD∥BE，??ADN??DBE．??DBE??BEM．
·························································· （2分） ?DB?DE，易得BE?2AD．得BE?8； ·
②当?ADB??BME时，?AD∥BE，??ADB??DBE． ??DBE??BME．又?BED??MEB，?△BED∽△MEB． DEBE1222222，即BE?EM?DE，得x???2?(x?4)?2?(x?4)．
BEEM2解得x1?2，x2??10（舍去）．即线段BE的长为2． ··········································· （2分） 综上所述，所求线段BE的长为8或2．
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