( 1999? Hangzhou) as shown in the figure, in △ABC, AM and BN intersect at D, BM=3MC, AD=DM, and find the value of: (1) BD: DN; (2) area S△ABN:
Solution: Solution: (1) Let c be the extension line of point E where CE∨AM intersects BA, and point F where BN intersects Ce.
∫CE∨AM，
∴∠DAN=∠FCN,∠ADN=∠CFN，
∴△DAN∽△FCN，
∴DNFN=ADCF，
And ∵AD=DM,
∴DNFN=DMCF，
∫CE∨AM，
∴BDBF=DMFC=BMBC=34，
∴DNFN=34，
∴BD:DN=3:37=7: 1.
(2) From (1):△ Dan is similar to △FCN.
∴ANCN=DNFN=34
∴S△ABN:S△CBN=AN:CN=3:4.