已知梯形ABCD中,CD//AB,∠ABC的角平分线BE⊥AD于E,且DE:AE=1:2,求S△ABE:S四边形BCDE

延长AD和BC交于F
∵BE是∠ABF(∠ABC)平分线
BE⊥AF(AD)
∴∠ABE=∠FBE,∠AEB=∠FEB=90°
∵BE=BE
∴△ABE≌△FBE(ASA)
∴AE=EF S△ABE=S△FBE1/2△ABF
DE∶AE=1∶2
DE=1/2AE
∵DE+DF=AF=AE
∴DF=1/2AE=DE
∴DF/AF=1/4
∵CD∥AB
∴△CDF∽△ABF
∴S△CDF/S△ABF=(DF/AF)²=(1/4)²=1/16
S△CDF=1/16S△ABF
∴S四边形BCDE
=S△FBE-S△CDF
=1/2S△ABF-1/16S△ABF=7/16S△ABF
∴S△ABE/S四边形BCDE=(1/2S△ABF)/(7/16S△ABF)=8/7
S△ABE:S四边形BCDE=8∶7