, 2008

Solution:
Since $D$ is the midpoint of the side $BC$, in the extension of the line segment $ZD$ we take a point $H$ such that $ZD = DH$. Then the quadrilateral $BHCZ$ is a parallelogram and therefore we have
$$BH = ZC = ZA$$

Also from the isosceles triangle $ABE$ we get
$$BE = AE$$
Since $DE \perp DZ$, $ED$ is altitude and median of the triangle $EZH$ and so this triangle is isosceles with
$$EH = EZ$$
From (1), (2) and (3) we conclude that the triangles $BEH$ and $AEZ$ are equal. Therefore they have also
$$\widehat{BEH} = \widehat{AEZ}, \quad \widehat{EBH} = \widehat{EAZ} \text{ and } \widehat{EHB} = \widehat{AZE}$$
Putting $\widehat{EBA} = \widehat{EAB} = \omega$, $\widehat{ZAC} = \widehat{ZCA} = \varphi$, then we have $\widehat{CBH} = \widehat{BCZ} = \widehat{C} + \varphi$, and therefore from the equality $\widehat{EBH} = \widehat{EAZ}$ we receive:
$$\begin{gathered} 360^{\circ} - \widehat{EBA} - \widehat{B} - \widehat{CBH} = \widehat{EAB} + \widehat{A} + \widehat{ZAC} \\ \Rightarrow 360^{\circ} - \widehat{B} - \omega - \varphi - \widehat{C} = \omega + \widehat{A} + \varphi \\ \Rightarrow 2(\omega + \varphi) = 360^{\circ} - (\widehat{A} + \widehat{B} + \widehat{C}) \\ \Rightarrow \omega + \varphi = 90^{\circ} \\ \Rightarrow \frac{180^{\circ} - \widehat{AEB}}{2} + \frac{180^{\circ} - \widehat{AZC}}{2} = 90^{\circ} \\ \Rightarrow \widehat{AEB} + \widehat{AZC} = 180^{\circ} \end{gathered}$$
From the supposition $EZ = 2 \cdot ED$, we get that the right triangle $ZEH$ has $\widehat{EZD} = 30^{\circ}$ and $\widehat{ZED} = 60^{\circ}$. Thus we have $\widehat{ZEH} = 120^{\circ}$.
However, since we have proved that $\widehat{BEH} = \widehat{AEZ}$, we get that
$$\widehat{AEB} = \widehat{AEZ} + \widehat{ZEB} = \widehat{ZEB} + \widehat{BEH} = \widehat{ZEH} = 120^{\circ}$$
From (5) and (6) we obtain that $\widehat{AZC} = 60^{\circ}$ and thus $\widehat{AEB} = 2 \cdot \widehat{AZC}$.