如图,△ABC中,AD、BE相交于点O,BD:CD=3:2,AE:CE=2:1.那么S△BOC:S△AOC:S△AOB为( ) A.2:3:4 B.2:3:5 C.3:4:5 D.3:4:6
问题描述:
如图,△ABC中,AD、BE相交于点O,BD:CD=3:2,AE:CE=2:1.那么S△BOC:S△AOC:S△AOB为( )
A. 2:3:4
B. 2:3:5
C. 3:4:5
D. 3:4:6
答
∵BD:CD=3:2,AE:CE=2:1,
∴S△AOB:S△AOC=3:2=6:4,S△AOB:S△COB=2:1=6:3,
∴S△BOC:S△AOC:S△AOB=3:4:6.
故选D.