Suppose that the line $A A^{\prime}$ intersects the lines $E Z, B C$ and $C N$ at the points $L, M$, $F$ respectively. The line $I K$ being diagonal of the rectangle $K A^{\prime} I A$ passes through $L$, which by construction of $A^{\prime}$, is the middle of the other diagonal $A A^{\prime}$. The triangles $Z A L, A L E$ are similar, so $\angle Z A L=\angle A E Z$. By the similarity of the triangles $A B C, D A B$, we get $\angle A C B=\angle B A D$. We have also that $\angle A E Z=\angle B A D$, therefore

$$\angle Z A L=\angle C A M=\angle A C B=\angle A C M$$

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Since $A F \perp C N$, we have that the right triangles $A F C$ and $C D A$ are equal. Thus the altitudes from the vertices $F, D$ of the triangles $A F C, C D A$ respectively are equal. It follows that $F D \| A C$ and since $D E \| A C$ we get that the points $E, D, F$ are collinear.

In the triangle $L F T$ we have, $A^{\prime} I \| F T$ and $\angle L A^{\prime} I=\angle L I A^{\prime}$, so $\angle L F T=\angle L T F$. Therefore the points $F, A^{\prime}, I, T$ belong to the same circle. Also, $\angle A^{\prime} I N=\angle A^{\prime} F N=90^{\circ}$ so the quadrilateral $I A^{\prime} F N$ is cyclic. Thus, the points $F, A^{\prime}, I, T, N$ all lie on a circle. From the above, we infer that

$$\angle N A^{\prime} T=\angle T F N=\angle A C F=\angle F E Z=\angle A D T \text {. }$$