Question 4
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 $BC$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that $$
AD + DT + TX + XA = CD + DY + YZ + ZC.
$$
AD + DT + TX + XA = CD + DY + YZ + ZC.
$$