解:作DM平行BC,交AE于M,则:∠DMF=∠CEF;又DF=CF;∠DFM=∠CFE.∴⊿DMF≌⊿CEF(AAS),DM=CE;MF=EF.∴DM/BE=CE/BE=CE/(2CE)=1/2.则AM/AE=AD/AB=DM/BE=1/2, AD=BD;AM=EM=2EF,AE=4EF.∴S⊿BDC=S⊿ADC=1/2.(等底同高的三角形面积相等)同理:S⊿CEA/S⊿BEA=CE/EB=1/2,则S⊿CEF=(1/3)S⊿ABC=1/3.又S⊿CEF/S⊿CEA=EF/AE=1/4,则S⊿CEF=(1/4)S⊿CEA=1/12.所以,S阴影=S⊿BDC-S⊿CEF=1/2-1/12=5/12....
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