..4分\r【拓展延伸】’."8=6,/£)=4,AG=2,AE=3,\r.AB_6_3AE_3.AB_AE\r**AD-4-2fAG-2'*AD~AG\rV四边形ABCD和AEFG是矩形,ZBAD=/EAG=90°,\r:./BAD+/DAE=ZEAG+ZDAE.即/BAE=ZDAG.......................................6分\r:.ABAEsADAG,B^^AB^3,NABE=NADG,.........................................7分\rDGAD2\r设AD与BE交于点P,BE与DG交于点、O,\r■:NDPE=ZAPB,NABE+NAPB=9Q°,AZADG+ZDPE=90°\r;.NDOB=90°.J.BEVDG,.................................................9分\r综上所述,线段BE和DG之间的关系为EE=3,BELDG....................................10分\rDG2\r25.解(I)正确比例式是:垣=殁,.................................1分\rBCAB\r.aE=AD-BC一(AB-BD)智一[\r=3分\r,,ABAB84....................\rA\r(2)另一个错误是没有进行分类讨论,如图,过点。作\r又=则△NDES/\/CB,DE=AD,....4分\rBCAC\rBC\r.•.。£=处呢="=2......................6分\rAC42\r综合以上可得:OE为:或擀.......................7分\r(3)当r=3或f=」2时以4,M,N为顶点的三角形与△28相似...........10分\r25
全文预览