长AC交直线l于点G. ………………………………………………(1分)Р∵∠ACB = 90°,∠ABC =∠GBC,∴∠BAC =∠BGC.Р∴AB = GB.…………………………………………………………………………………(1分)Р∴AC = GC.…………………………………………………………………………………(1分)Р∵AE⊥l,CD⊥l,∴AE∥CD.Р∴. …………………………………………………………………………(1分)Р∴AE = 2CD. ………………………………………………………………………………(1分)Р(3)(I)如图1,当点E在DB延长线上时:Р(第25题图1)РAРCРDРEРlРGРBРHРFР过点C作CG∥l交AB于点H,交AE于点G,则∠CBD =∠HCB.Р∵∠ABC =∠CBD,∴∠ABC =∠HCB.∴CH = BH.………(1分)Р∵∠ACB = 90°,∴∠ABC +∠BAC =∠HCB +∠HCA = 90°.Р∴∠BAC =∠HCA.∴CH = AH = BH.РBР(第25题图2)РAРCРDРlРGРEРHРFР∵CG∥l,∴.Р设CH = 5x,则BE = 6x,AB = 10x.Р在Rt△ABE中,.Р由(2)知AE = 2CD = 8,∴,得.Р∴CH = 5,BE = 6,AB = 10.Р∵CG∥l,∴,∴HG=3.……………………(1分)Р∴CG = CH + HG = 8.Р易证四边形CDEG是矩形,∴DE = CG = 8.Р∴.…………………………………………(1分)Р(II)如图2,当点E在DB上时:Р同理可得CH = 5,BE = 6,HG = 3.…………………………(1分)Р∴.Р∴BD=DE + BE = 8.…………………………………………………………………………(1分)Р综上所述,BD的长为2或8.
全文预览
