2.因为 △AEF∽△ACB,AF/AB=AE/AC=1/3故角AEF=角C,又角AEF=角DEB,故△DEB~△FDC。则有BE/FC=DB/DF=DE/DC,其中,BE=3,FC=7,DF=DE+EF=DE+BC/3=DE+4,DC=DB+BC=DB+12即3/7=DB/(DE+4)=DE/(DB+12),解得,BD= 4.8
DE=7.23.若BC=X,DE=Y,则3/7=Y/(DB+X),其中DB=(Y+X/3)*3/7=,整理得y=0.6x .............................
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