题目 #20603

【讲解】 ①当点在BC上时. D 如图连接,AC交MN于点O, QVABC是等边三角形, AB = = = 30,\(\frac{AC}{MN}\frac{BC}{折叠得到}\)A Q 直线将lY, 是AC的中点,\(\frac{ABCD}{四边形}\frac{的面积平分}{ABCD}\frac{O}{为平行四边形} =\) =  = 60o,QVAMN沿 V, B C\(DMN_{o}\)OA = OC; ,\(\frac{Q}{CD} \frac{AB}{NAO}\)所以AN = DN, AM =, = MDN = 60,在\(\frac{DM}{DNC}\)A P, = = 3,  = ,\(\frac{CDAB}{MCO} \frac{所以}{V}\frac{ANO}{MCOAAS} \frac{CMO}{AN}\)V 中, +  = + BMD,\(\frac{BMD}{BMD} \frac{MDN}{DNC} \frac{所以}{V}\)B = ,V  (), =;\(Q \frac{NAO}{EDM} \frac{CM}{ENA}\) = .又 Q  = ，VBMD  CDN, B C AB PCD, = EAN,EMD = ,所 \(\frac{BD}{CN} = \frac{BM}{DC} = \frac{MD}{DN}\). Q BC = 30, BD : DC =1: 4, 以VEDM VEAN,所以\(\frac{DM}{AN} = \frac{DE}{AE} = = \frac{1}{3}\)即 2 2 + 4 BD = 6, DC = 设AN = x,则DN =, = 30 −\(24. \frac{xCN}{144} 3DM =\)AN,3(3− CM) = CM,解得 : CM =\(\frac{9}{4}\)故 x, =\(\frac{BM}{24} = \frac{MD}{x},BM =\)MD = 6 30 − x − x 6x\(\frac{30}{BM}\)答案为:\(\frac{9}{4}\). 6x , =,因为AM + = 30,所以\(\frac{AM}{144} 30\)− x 30 − x 6x + = 30, x = 21, AN = 21; 30 − x 30 − x ②当点在BC的反向延长线上时. D 2026-06-22 与①同理可得VBMD VCDN,\(\frac{BD}{CN} = \frac{BM}{DC} = \frac{MD}{DN},Q\)BC = 30,且? : =1: 4,BD =10,\(\frac{BDDC}{DN}\)DC = 40,设AN = x,则 = x,CN = x − 30,所 以 =\(\frac{BM}{40} = \frac{MD}{x},BM = \frac{400}{30}\)MD = 10 − x −\(\frac{x}{10} \frac{30}{Q} \frac{x}{30}\)AM + = 30,\(\frac{x}{30} + \frac{400}{30} = 30, 10 \frac{BM}{AN}\)− x − x −\(\frac{x}{故答案为}\)解得x = 65, = : 21或 65. 65.