Question

已知:如图,AB平行于CD,AD与BC交于点F,AF=BF,点E在AB上,CE与AD交于O,EC=EB,OF=4,FD=

Answer 1

⑴∵AF=BF,∴∠A=∠B,

∵AB∥CD,∴∠B=∠BCD,∠A=∠D,

∵BE=CE,∴∠B=∠OCF,

∴∠OCF=∠D,又∠COD=∠COD,

∴ΔOCF∽ΔODC,∴OC/OF=OD/OC,

∴OC²=OF*OD.

⑵OC²=OF*OD=36,∴OC=6,

∵∠BCD=∠D,∴CF=DF=5,

由⑴相似得：CF/CD=OF/OC=2/3,

∴CD=3/2×5=15/2.

⑶∵OE：OC=2/3,∴OE=4,

∵AB∥CD,∴OA/OD=OE/OC,OA=6,

∴SΔABF/SΔCDF=(6/5)²=36/25,

SΔCDF=25/36*SΔABF,

又SΔOCF/SΔCDF=OF/DF=4/5(同高),

∴SΔOCF=4/5SΔCDF=5/9SΔABF,

∴SΔOCF：SΔABF=5：9.