△ABC中,作直线DN平行于中线AM,设这条直线交边AB于D,交边CA的延长线于E,交边BC于N.证:AD:AB=AE:AC

问题描述:

△ABC中,作直线DN平行于中线AM,设这条直线交边AB于D,交边CA的延长线于E,交边BC于N.证:AD:AB=AE:AC

证明:
AD:AB = MN:MB |
| => AD:AB=MN:MC |
MB=MC | | => AD:AB=AE:AC
MN:MC=AE:AC |
.

因为AM||EN,所以AD:AB=MN:MB
在CA的延长线上做FB||AM,所以MN:MB=AE||AF
又因为M为BC中点,CA:AF=CM:MB=1,即AC=AF
所以AD:AB=AE:AC

DN//AM=> AD:AB=NM:BM
因为:在三角形ABCK AM是中线=》BM=CM
所以:AD:AB=NM:CM
因为:AM//DN=> AM//EN
所以:AE:AC=MN:CM
综合以上两个所以得出:AD:AB=AE:AC

AD:AB=MN:MB
MB=MC
AD:AB=MN:MC=AE:AC