First, let us denote $\angle B=\alpha, AD\cap EZ={P}, IK\cap AA'={R}$ and $AA'\cap DE={Q}$. Since $DE||AC$, we have $\angle DEA=90^o$. Similarly, $\angle DZA=90^o$. Now, since $\angle ADB=90^o$, we have $\angle DAB = 180^o-\angle ADB -\angle ABD=180^o-90^o-\alpha=90^o-\alpha$. From here, $\angle ADE=180^o-\angle DEA-\angle DAE=180^o-90^o-(90^o - \alpha) = \alpha$. Because $AZDE$ is a rectangle, $\angle ADE=\angle AZE \equiv \angle AZR$. $ZE$ and $CN$ are parallel, so $\angle AZE=\angle ACN=\alpha$. From here we derive $\angle CNA=90^o-\angle ACN=90^o-\alpha$ and $\angle INC=180^o-\angle CNA=90^o+\alpha$ $A'$ is symmetric of $A$ with respect to $ZE$, so we must have $AA'\perp ZE$. Thus, $\angle ZAR=90^o-\angle AZR=90^o-\alpha$. Now, $\angle AA'K=90^o-\angle KAA' \equiv 90^o-\angle ZAR=\alpha$ Because $AKA'I$ is a rectangle, we get$\angle AA'K=\angle AIK=\alpha$ So, $\angle ETI=90^o-\angle EIT \equiv 90^o-\angle AIK=90^o-\alpha$ and $\angle ITQ=180^o-\angle ETI=90^o+\alpha$ Lemma 1: $CN, AA'$ and $DE$ are concurrent Proof: 1) Let us suppose a line $p$ through $Q$, such that $p\perp AA'$. Let $C'$ be the intersection of $p$ with $AC$. Since $\angle CAA'=90^o-\alpha$, and therefore $\angle C'AA'=90^o-\alpha$, $\angle QC'A=\alpha$ Because of this, and because $\angle ADQ \equiv \angle ADE=\alpha$, $\angle QC'A=\angle QDA, C'AQD$ is cyclic, a.e. the excircle of $\triangle AQD$ contains $C'$. 2) We have $\angle A'AI \equiv \angle QAE=\alpha$. Since $\angle DEA \equiv \angle QEA=90^o$, we obtain $\angle DQA=\angle QAE+\angle QEA=90^o+\alpha$ By definition, $\angle ACB \equiv \angle DCA=90^o-\alpha$. Now we see that $\angle DCA+\angle DQA=90^o-\alpha+90^o+\alpha=180^o$, so the excircle of $\triangle AQD$ must contain $C$. Combining 1) and 2), we see that the excircle of $\triangle AQD$ intersects the line $AC$ at three points: $A, C$ and $C'$. It is impossible for all three to be distinct points, since then the excircle of $\triangle AQD$ will have three common points with $AC$, which is impossible. It follows that two of $A, C$ and $C'$ must be the same point. $A$ and $C'$ cannot be equivalent because, in that case $QC$' couldn't be perpendicular to $AA'$. So $C\equiv C'$. Now, $CQ$ is perpendicular to $AA'$, which means that $CQ\parallel ZE$, and that $CQ$ contains $N$. Thus, the proof of the lemma is complete. Lemma 2: $A'QNI$ is cyclic Proof: We have $\angle A'QN=90^o$ and $\angle A'IN=90^o$. Case 1: $N$ is between $A$ and $I$ -- $\angle A'IN+\angle A'QN=180^o$, so the lemma is true. Case 2: $N$ is between $B$ and $I$ -- $\angle A'QN=\angle A'IN=90^o$, so the lemma is true. So, $\angle CNI \equiv \angle QNI=\angle DTI \equiv \angle QTI$, or $\angle QTI=\angle QNI$. From here we obtain that $QTNI$ is cyclic. Because $QNIA'$ is also cyclic, $Q, T, N, I$ and $A'$ all lie on the same circle. *Note: When $N$ is between $I$ and $B$, the proof is done by using the fact that $\angle QNI \equiv\angle CNA=90^o -\alpha$ Case 1: $N$ is between $A$ and $I$ -- $\angle AIK \equiv \angle NIT=\alpha$, and by peripheral angles, we obtain $\angle NA'T=\angle NIT=\alpha$, a.e. $\angle NA'T=\angle ADT=\alpha$ . Thus, the proof is complete. Case 2: $N$ is between $B$ and $I$ -- $\angle NA'T=180^o-\angle TIN=\alpha$, so $\angle NA'T=\alpha=\angle ADT$. Thus, the proof is complete. *Note: When $N$ and $I$ are equivalent, the proof follows directly from the fact that $QNIA'$ is cyclic and $\angle CQD=\alpha$ .

Let $ED$ and $AA'$ meet $CN$ at $X_1$ and $X_2$ respectively. Note that \[ \frac{NX_1}{X_1C} = \frac{NE}{EA} = \frac{CZ}{ZA} = \frac{CD}{DB} = \frac{CA^2}{AB^2} = \frac{NA^2}{AC^2} = \frac{NX_2}{X_2C}. \]Thus $X_1=X_2 := X$. Easily check that $TIA'X$ is an isosceles trapezoid. Since $\angle A'XN=90^{\circ} = \angle NIA'$ then $NIA'X$ is cyclic. We conclue that $NTIA'X$ is cyclic and thus $\angle NA'T = \angle NIT = \angle ADT$.