Question 12
Let $AD$ be the altitude of the right angled triangle $ABC$ with $\angle A = 90^{\circ}$. Let $DE$ be the altitude of the triangle $ADB$ and $DZ$ be the altitude of the triangle $ADC$ respectively. Let $N$ be chosen on the line $AB$ such that $CN$ is parallel to $EZ$. Let $A'$ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. Prove that $\angle NA'T = \angle ADT$, where $T$ is the intersection point of $IK$ and $DE$.