过E作EP⊥AC交AC于P,由AE平分⊥BAC,∴BE=EP.设AB=1,AC=√2,设BE=EP=x,CE=x√2=1-x,由x√2=1-x,2x²=1-2x+x²x²+2x-1=0,∴x=-1±√2,将x=-1-√2<0舍去,∴x=√2-1,由OF‖PE,∴OF/AO=PE/AP,OF/(√2/2)/(√2-1)/(√2-x)OF=(2-√2)/2,EC=x√2=(√2-1)√2=2-√2.∴EC=2OF.证毕,
如图,正方形abcd的对角线ac,bd交于o,∠bac的平分线bd交于f,交bc于e,求证,CE=2OF
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