Let $\Gamma$ be a circle with centre $I$, 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

Step 1 (Tangent lengths): Since $ABCD$ is tangential to $\Gamma$ with center $I$: $AB+CD=BC+DA$ (Pitot's theorem). Let the tangent lengths from each vertex be: $AP=AS=p$, $BP=BQ=q$, $CQ=CR=r$, $DR=DS=s$, where $P,Q,R,S$ are tangent points on $AB,BC,CD,DA$.

Step 2 (Properties of $\Omega$): $\Omega$ is the circumcircle of $△AIC$. Since $I$ is the incenter of the tangential quadrilateral, $\angle AIC=90^{\circ }+\angle B/2$ (for a tangential polygon, this follows from $\angle AIB+\angle CID=180^{\circ }$ etc.).

Step 3 (Points $X,Z,Y,T$): $X=BA\cap \Omega$ (beyond $A$), $Z=BC\cap \Omega$ (beyond $C$), $Y=AD\cap \Omega$ (beyond $D$), $T=CD\cap \Omega$ (beyond $D$).

Step 4 (Power of point $D$): Power of $D$ w.r.t. $\Omega$: $DA\cdot DY=DC\cdot DT$ (since $Y\in \Omega$ on ray $DA$ and $T\in \Omega$ on ray $DC$). So $DY=\frac{DC\cdot DT}{DA}$.

Step 5 (Compute each sum): LHS $=AD+DT+TX+XA$, RHS $=CD+DY+YZ+ZC$. Using the tangent length relations and the power-of-a-point identities, express $TX,XA,DY,YZ$ in terms of the tangent lengths. The equality reduces to an identity involving $p,q,r,s$.

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