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∴60NEFq=?, ∴DE为NEF的平分线. 从而D到EN与EF的距离相等,即J到CF与EF的距离也相等. 又因为FCJECJ=?所以J为△CEJ的内心. 证三 如图,建立 直角坐标系,依题意可 设(cos2,sin2)Aqq、 (cos,sin)Dqq、 (cos(260),Fq+皊in(260))q+? 由D是弧AB的中点,知//ACOD. ∴sin2sincos2cosOJD
∴60NEFq=?, ∴DE is the NEF bisector. Thus the distance from D to EN and EF is equal, that is, the distance from J to CF and EF is also equal. And because FCJECJ=?, so J is the heart of △CEJ. Figure, the establishment of a rectangular coordinate system, according to the title can be set (cos2, sin2) Aqq, (cos, sin) Dqq, (cos (260), Fq + 皊in (260)) q +? By D is the midpoint of the arc AB, Know //ACOD. ∴sin2sincos2cosOJD