的中点,
∴AE=BE,∴∠FBE=∠BCE又∵∠FEB=∠BEC,∴△EBF∽△ECB,∴EF∶BE=BE∶EC,∴EFEC=BE2=23r2=49r23如解图②,连结OA
∵AE=BE,
∴AE=BE=23r设OH=x,则HE=r-x,在Rt△OAH中,AH2+OH2=OA2,即AH2+x2=r2,在Rt△EAH中,AH2+EH2=EA2,即AH2+r-x2=23r2,解得x=79r,∴HE=r-79r=29r在Rt△OAH中,AH=OA2-OH2=492r∵OE⊥AB,∴AH=BH又∵F是AB的四等分点,∴HF=12AH=292r在Rt△EFH中,EF=HE2+HF2=293r∵EFEC=49r2,∴293rEC=49r2,
6
f∴EC=233r
7
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