Let $AD$ be the perpendicular to the hypotenuse $BC$ of the right triangle $ABC$. Let $DE$ be the height of the triangle $ADB$ and $DZ$ be the height of the triangle $ADC$. On the line $AB$ is chosen the point $N$ so that $CN$ is parallel to $EZ$. Let $A'$ be symmetrical of $A$ to $EZ$ and $I, K$ projections of $A'$ on $AB$, respectively, on $AC$. Prove that $<$ $NA'T$ $=$ $<$ $ADT$, where $T$ is the point of intersection of $IK$ and $DE$.