∵ac:cb=2:5,ad:db=5:6∴ac=2/7ab,cb=5/7ab；ad=5/11ab,bd=6/11ab∵ad=ac+cd,cd=13∴5/11ab=2/7ab+13∵13=13/77ab∴ab=77注：你是初二的吧？那应该还没学二元一次方程，这是解线段题的标准写法。
设ab=x则ac=2/7x cb=5/7x ad=5/11x bd=6/11xcd=ad-ac=cd-bd=13没带笔 自己算吧
设 ac = x, cd = y, db = z.x/z= 2/5(x+y)/z = 5/6y = 13求得x = 12z = 30ab = x+y+z = 12+13+30 = 55
ac/cb=2:5(ac+cd)/(cb-cd)=5:6(ac+13)/(cb-13)=5:6cb=55ac=22ab=77
答：解:点C为AB的三等分点,可能是AC=(1/3)AB或BC=(1/3)AB. 1)当AC=(1/3)AB时(如上图): MN=AN-AM=(1/2)AB-AC-...详情>>
