JBMO-SL 2018 G2

Let $A B C$ be a right angled triangle with $\angle A=90^{\circ}$ and $A D$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $A B$ and $A C$ respectively. The parallel line from $C$ to $E Z$ intersects the line $A B$ at the point $N$ Let $A^{\prime}$ be the symmetric of $A$ with respect to the line $E Z$ and $I, K$ the projections of $A^{\prime}$ onto $A B$ and $A C$ respectively. If $T$ is the point of intersection of the lines $I K$ and $D E$, prove that $\angle N A^{\prime} T=\angle A D T$.