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    B Proof of Cyclic Quadrilateral AEDT in Circle ABCD

    I've just known that ##\angle TEA = \angle TDA## prove ##AEDT## are cyclic. The problem is solved. Thanks.
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    B Proof of Cyclic Quadrilateral AEDT in Circle ABCD

    ##TA## and ##TD## are tangent line of circle ##ABCD## and ##TB \parallel DC##. Show ##A,E,D,T## are cyclic quadrilateral. I know ##x=\angle TAD= \angle TDA = \angle ACD= \angle TEA## And ##\angle ATD=180-2x## But I don't know how to prove ##\angle AED=x##. Or there's another easily method? Thanks.
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