如图,在△ABC中,BF交AD于E. (1)若AE:ED=2:3,BD:DC=3:2,求AF:FC; (2)若AF:F
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![如图,在△ABC中,BF交AD于E. (1)若AE:ED=2:3,BD:DC=3:2,求AF:FC; (2)若AF:F](/uploads/image/z/5570668-28-8.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CBF%E4%BA%A4AD%E4%BA%8EE.+%281%29%E8%8B%A5AE%3AED%3D2%3A3%2CBD%3ADC%3D3%3A2%2C%E6%B1%82AF%3AFC%EF%BC%9B+%282%29%E8%8B%A5AF%3AF)
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如图,在△ABC中,BF交AD于E. (1)若AE:ED=2:3,BD:DC=3:2,求AF:FC; (2)若AF:F
如图,在△ABC中,BF交AD于E. (1)若AE:ED=2:3,BD:DC=3:2,求AF:FC; (2)若AF:F
如图,在△ABC中,BF交AD于E. (1)若AE:ED=2:3,BD:DC=3:2,求AF:FC; (2)若AF:F
1、过点D作DG∥BF交AC于G,设FG=3K
∵DG∥BF
∴AE/ED=AF/FG
∵AE:ED=2:3
∴AF/FG=2/3
∴AF/3K=2/3
∴AF=2K
又∵DG∥BF
∴FG/CG=BD/CD
∵BD:CD=3:2
∴FG/CG=3/2
∴3K/CG=3/2
∴CG=2K
∴FC=FG+CG=3K+2K=5K
∴AF/FC=2K/5K=2/5
∴AF:FC=2:5
2、过点D作DG∥BF交AC于G,设FG=4K
∵DG∥BF
∴BD/CD=FG/CG
∵BD:CD=4:3
∴FG/CG=4/3
∴4K/CG=4/3
∴CG=3K
∴CF=FG+CG=4K+3K=7K
∵AF:FC=2:7
∴AF=2K
又∵DG∥BF
∴AE/ED=AF/FG=2K/4K=1/2
∴AE:ED=1:2
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