在平行四边形ABCD中，E、F是对角线BD上的点，且BE=EF=FD，连接AE交BC于点M，连接MF交AD于点H，则△AMH和平行四边形ABCD的面积比为
3:8
分析：由平行四边形的性质及平行线的性质求出AH：AD的值，再根据△AMH与?ABCD等高，利用面积公式求底边的比．
解：∵BE=EF=FD，
∴DE=2BE，BF=2DF，
∵AD∥BC，
∴△ADE∽△MBE，△BMF∽△DHF，
∴
，即BM=
AD，
同理可得DH=BM=
AD，
∴AH=AD-DH=
AD，
设△AMH的AH边上高为h，
则
．
故答案为：3：8．
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