Suppose that the line $AA'$ intersects the lines $EZ$, $BC$ and $CN$ at the points $L$, $M$, $F$ respectively. The line $IK$ being diagonal of the rectangle $KA'IA$ passes through $L$, which by construction of $A'$ is the middle of the other diagonal $AA'$. The triangles $ZAL$ and $ALE$ are similar, so $\angle ZAL = \angle AEZ$. By the similarity of the triangles $ABC$ and $DAB$ we get $\angle ACB = \angle BAD$. We have also that $\angle AEZ = \angle BAD$, therefore
$$\angle ZAL = \angle CAM = \angle ACB = \angle ACM$$
Since $AF \perp CN$, we have that the right angled triangles $AFC$ and $CDA$ are equal. Thus the altitudes from the vertices $F$ and $D$ of triangles $AFC$ and $CDA$ are respectively equal. It follows that $FD \parallel AC$ and since $DE \parallel AC$ we get that the points $E, D, F$ are collinear. In triangle $LFT$ we have
$$A'I \parallel FT \text{ and } \angle LA'I = \angle LIA',$$
so $\angle LFT = \angle LTF$. Therefore the points $F, A', I, T$ belong to the same circle. Also $\angle A'IN = \angle A'FN = 90^{\circ}$ so the quadrilateral $IA'FN$ is cyclic. Thus the points $F, A', I, N$ all lie on a circle. From the above, we infer that
$$\angle NA'T = \angle TFN = \angle ACF = \angle FEZ = \angle ADT$$