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We denote $P= WX \cap YZ$ to mean point $P$ is the intersection of lines $WX$ and $YZ$.

The problem is about pascal's theorem: Let $ABCD$ be a cyclic quadrilateral. Let the tangent lines at A and at B to the circumcircle of ABCD meet at R. Let the tangent lines at C and at D to the circumcircle of ABCD meet at S. Let $T=AD \cap BC$ and $U=AC \cap BD$. Prove that $R,S,T,U$ are collinear.

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up vote 2 down vote accepted

Considering $(A,A,C,B,B,D)$, what do you get?

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we have T,U and R are collinear. Then we apply Pascal's theorem again to $(A,D,D,B,C,C)$ to get T,U,S collinear? Thank you very much! – Ishigami May 5 '13 at 18:17

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