Since $D$ is the midpoint of the side $B C$, in the extension of the line segment $Z D$ we take a point $H$ such that $Z D=D H$. Then the quadrilateral $B H C Z$ is a parallelogram and therefore we have

$$B H=Z C=Z A$$

!

Also from the isosceles triangle $A B E$ we get

$$B E=A E$$

Since $D E \perp D Z, E D$ is the altitude and median of the triangle $E Z H$ and so this triangle is isosceles with

$$E H=E Z$$

From (1), (2) and (3) we conclude that the triangles $B E H$ and $A E Z$ are equal. Therefore they have also

$$\widehat{B E H}=\widehat{A E Z}, \widehat{E B H}=\widehat{E A Z} \text { and } \widehat{E H B}=\widehat{A Z E}$$

Putting $\widehat{E B A}=\widehat{E A B}=\omega, \widehat{Z A C}=\widehat{Z C A}=\varphi$, then we have $\widehat{C B H}=\widehat{B C Z}=\widehat{C}+\varphi$, and therefore from the equality $\widehat{E B H}=\widehat{E A Z}$ we receive:

$$\begin{gathered} 360^{\circ}-\widehat{E B A}-\widehat{B}-\widehat{C B H}=\widehat{E A B}+\widehat{A}+\widehat{Z A C} \\ \Rightarrow 360^{\circ}-\widehat{B}-\omega-\varphi-\widehat{C}=\omega+\widehat{A}+\varphi \\ \Rightarrow 2(\omega+\varphi)=360^{\circ}-(\widehat{A}+\widehat{B}+\widehat{C}) \\ \Rightarrow \omega+\varphi=90^{\circ} \\ \Rightarrow \frac{180^{\circ}-\widehat{A E B}}{2}+\frac{180^{\circ}-\widehat{A Z C}}{2}=90^{\circ} \\ \Rightarrow \widehat{A E B}+\widehat{A Z C}=180^{\circ} \end{gathered}$$

From the supposition $E Z=2 \cdot E D$, we get that the right triangle $Z E H$ has $\widehat{E Z D}=30^{\circ}$ and $\widehat{Z E D}=60^{\circ}$. Thus we have $\widehat{Z E H}=120^{\circ}$.

However, since we have proved that $\widehat{B E H}=\widehat{A E Z}$, we get that

$$\widehat{A E B}=\widehat{A E Z}+\widehat{Z E B}=\widehat{Z E B}+\widehat{B E H}=\widehat{Z E H}=120^{\circ}$$

From (5) and (6) we obtain that $\widehat{A Z C}=60^{\circ}$ and thus $\widehat{A E B}=2 \cdot \widehat{A Z C}$.