Iranian Mathematical Olympiad

Let $I$ be the incenter and $I_a$ the $A$-excenter of triangle $ABC$ and $J$ the $C$-excenter of triangle $CDE$. $J$ lies on $\omega$, since $$\angle EJD=90^{\circ }-\frac{1}{2}\angle ECD=\angle CDE.$$ We claim that $A$, $J$, and $Z$ are collinear. Point $T$ lies on $AC$ such that $IT\parallel DE$. To prove our claim, we just need to show that $\frac{IJ}{IT}=\frac{I_{a}Z}{I_{a}C}$, since $IT\parallel C{I}_a$, $IJ\parallel I_{a}Z$ and $I{I}_a$ passes through $A$. Notice that $$\angle ITE=\angle DEC=\angle EDC=\angle I_{a}CY⟹△IET\sim △I_{a}YC⟹\frac{IE}{IT}=\frac{I_{a}Y}{I_{a}C},$$ hence $\frac{IJ}{IT}=\frac{I_{a}Z}{I_{a}C}$, since $IE=IJ$ and $I_{a}Y=I_{a}Z$. So the claim is proved. Let $BC$ touches $A$-excircle at $Y$. The extensions of sides of quadrilateral $AEDB$ are tangent to $A$-excircle, therefore $$AE+ED=AB+BD⟹ED=2BD.(1)$$ Yielding $DX=DE$, and it is obvious that $DY=DZ$ so quadrilateral $YZXE$ is an isosceles trapezoid and $Y$ lies on the circumcircle of triangle $XZE$. Hence $$\angle EKY=\angle EZY=\frac{1}{2}\angle EDY=\angle EDM,$$ where $M$ is the midpoint of arc $ED$ (not containing $J$), so $KY$ passes through $M$. Let $N$ be the midpoint of segment $ED$. Notice that $△CDI\sim △CND$, hence $$\frac{DN}{CD}=\frac{ID}{IC}\stackrel{(1)}{⟹}\frac{BD}{CD}=\frac{IJ}{IC}⟹JB\parallel ID.$$

therefore $\angle YJD=\angle BKD$, since $\angle BJY=\angle BKY$. According to our assumption, $J$ lies on the angle bisector of $\angle YDZ$ and $YD=DZ$. It yields that $JD$ is the perpendicular bisector of $YZ$ and $\angle YJD=\angle ZJD$. Hence $\angle ZJD=\angle BKD$ and the result follows.