SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Suppose that the line $A{A}^{\prime }$ intersects the lines $EZ$, $BC$ and $CN$ at the points $L$, $M$, $F$ respectively. The line $IK$ being diagonal of the rectangle $K{A}^{\prime }IA$ passes through $L$, which by construction of $A^{\prime }$ is the middle of the other diagonal $A{A}^{\prime }$. 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 $$A^{\prime }I\parallel FT\text{ and }\angle L{A}^{\prime }I=\angle LI{A}^{\prime },$$ so $\angle LFT=\angle LTF$. Therefore the points $F,A^{\prime },I,T$ belong to the same circle. Also $\angle A^{\prime }IN=\angle A^{\prime }FN=90^{\circ }$ so the quadrilateral $I{A}^{\prime }FN$ is cyclic. Thus the points $F,A^{\prime },I,N$ all lie on a circle. From the above, we infer that $$\angle N{A}^{\prime }T=\angle TFN=\angle ACF=\angle FEZ=\angle ADT$$