Junior Balkan Mathematical Olympiad Shortlist

Solution:

In $\triangle ABD$ the ray $[AZ$ bisects the angle $\widehat{DAB}$ and the line $ZE$ is the perpendicular bisector of the side $[BD]$. Hence $Z$ belongs to the circumcircle of $\triangle ABD$.
Therefore the points $A, B, D, Z$ are concyclic.
In $\triangle BCD$, $AE$ and $ZE$ are the perpendicular bisectors $[BC]$ and $[BD]$, respectively. Hence, $E$ is the circumcenter of $\triangle BCD$ and therefore $\widehat{DEZ} = \widehat{BED} / 2 = \widehat{ACB}$. Since $BD \perp ZE$, we conclude that: $\widehat{BDE} = 90^{\circ} - \widehat{DEZ} = 90^{\circ} - \widehat{ACB} = \widehat{BAE}$. Hence the quadrilateral $A E B D$ is cyclic, that is the points $A, B, D, E$ are concyclic. Therefore, since $A, B, D, Z$ are also concyclic, we conclude that $A E Z D$ is cyclic.

Solution:

In $\triangle ABD$ the ray $[AZ$ bisects the angle $\widehat{DAB}$ and the line $ZE$ is the perpendicular bisector of the side $[BD]$. Hence $Z$ belongs to the circumcircle of $\triangle ABD$.
Therefore the points $A, B, D, Z$ are concyclic.
Let $M$ and $N$ be the midpoints of the sides $[BC]$ and $[DB]$, respectively. Then $N \in ZE$ and $M \in AE$. Next, $[MN]$ is a midline in $\triangle BCD$, so $MN \parallel CD \Rightarrow \widehat{NMB} \equiv \widehat{ACB}$.
But $[AZ$ is the external bisector of the angle $\widehat{BAC}$ of $\triangle ABC$, hence $\widehat{BAZ} \equiv \widehat{ACB}$. Therefore, $\widehat{NMB} \equiv \widehat{BAZ}$. In the quadrilateral $BMEN$ we have $\widehat{BNE} = \widehat{BME} = 90^{\circ}$, so $BMEN$ is cyclic $\Rightarrow \widehat{NMB} \equiv \widehat{BEZ}$, hence $\widehat{BAZ} \equiv \widehat{BEZ} \Rightarrow AEBZ$ is cyclic.
Therefore, since $A, B, D, Z$ are also concyclic, we conclude that $A E Z D$ is cyclic.

Solution:

Let $T$ be the symmetric of $B$ with respect to the axis $AZ$. Obviously $T \in AD$. Since $AE$ and $BT$ are both perpendiculars to $AZ$, they are parallel, so $\widehat{BAC} \equiv \widehat{BTA}$. (1)
Since $ZB = ZT = ZD$, the point $Z$ is the circumcenter of $\triangle BDT$.
Therefore $\widehat{BTA} = \widehat{BZD} / 2 = \widehat{EZD}$.
From (1) and (2) we conclude that $\widehat{EAC} \equiv \widehat{EZD}$, which gives that $AEDZ$ is cyclic.