Question

如图,BD是平行四边形ABCD的对角线,BE=EF=FD,则S△AMH:S△ABCD=____.

Answer 1

解:AD平行BC,则:DH:BM=DF:BF=1:2,BM=2DH;

AD:BM=DE:BE=2:1,即AD=2BM=4DH.

故AH=3DH.设三角形AMH中AH边上的高为h,则:

S△AMH:S△ABCD=(AH*h/2):(AD*h)=(3DH*h/2):(4DH*h)=3:8.