Let BD, EF and tagent at C concur at G Since tagent at C and EF concur at G so by some sin calculations we have $\dfrac{GE}{GF} = \dfrac{CE^2}{CF^2}$ According to Menelaus theorem, $\triangle$ ABD has F,E,G collinear so $\dfrac{GE}{GF} = \dfrac{BE}{BA}. \dfrac{DA}{DF}$ so $\dfrac{BE}{BA}. \dfrac{DA}{DF} = \dfrac{CE^2}{CF^2}$ but $\dfrac{AD}{BA} = \dfrac{AD}{DC} = \dfrac{sinDCA}{sinDAC} = \dfrac{sinCFE}{sinCEF} = \dfrac{CE}{CF}$ so that $\dfrac{CE}{CF} = \dfrac{CB}{CD} = \dfrac{BE}{DF}$ which mean $\triangle CFD \sim \triangle CEB$ $\angle CFD = \angle CEA = \angle CEB$ but $\angle CEA + \angle CEB = 180^{o}$ so $ \angle CEA = \angle CFA = 90^o$ <=> AC is diameter of k

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