如图,在△ABC中,M是AC的中点,P,Q为边BC的三等分点.若BM与AP,AQ分别交于D,E两点,则BD,DE,EM三条

2025-01-21 15:45:15
推荐回答(1个)
回答1:

过A作AF BC交BM延长线于F,设BC=3a
则BP=PQ=QC=a;
∵AM=CM,AF BC,
∴AF:BC=AM:CM=1,
∴AF=BC=3a,
∴BD:DF=BP:AF=1:3,
∴BD=
BF
4

同理可得:
BE=
2BF
5
,BM=
BF
2

∴DE=BE-BD=
3BF
20
,EM=BM-BE=
BF
10

∴BF:FG:GE
=
1
4
3
20
1
10

=5:3:2;
故选C.