Discussion on International Mathematical Olympiad (IMO)
Let $\Gamma$ be a circle with center $I$, and $ABCD$ a convex quadrilateral such that each of the segments $AB, BC, CD$, and $DA$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $AIC$. The extension of $BA$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[AD + DT + TX + XA = CD + DY + YZ + ZC.\]
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