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 \[ BH = ZC = ZA. \tag{1} \] Also, from the isosceles triangle \( ABE \), we get \[ BE = AE. \tag{2} \] Since \( DE \perp DZ \), \( ED \) is the altitude and median of the triangle \( EZH \), and so this triangle is isosceles with \[ EH = EZ. \tag{3} \] From (1), (2), and (3), we conclude that the triangles \( BEH \) and \( AEZ \) are equal. Therefore, they also have \[ \angle BEH = \angle AEZ, \quad \angle EBH = \angle EAZ, \quad \text{and} \quad \angle EHB = \angle AZE. \tag{4} \] Putting \( \angle EBA = \angle EAB = \omega \), and \( \angle ZAC = \angle ZCA = \varphi \), then we have \( \angle CBH = \angle BCZ = \text{C} + \varphi \), and therefore, from the equality \( \angle EBH = \angle EAZ \), we get: \[ 360^\circ - \angle EBA - \text{B} - \angle CBH = \angle EAB + \text{A} + \angle ZAC \]\[ \Rightarrow 360^\circ - \text{B} - \omega - \varphi - \text{C} = \omega + \text{A} + \varphi \]\[ \Rightarrow 2(\omega + \varphi) = 360^\circ - \left(\text{A} + \text{B} + \text{C}\right) \]\[ \Rightarrow \omega + \varphi = 90^\circ \]\[ \frac{180^\circ - \angle AEB}{2} + \frac{180^\circ - \angle AZC}{2} = 90^\circ \] \] \[ \Rightarrow \angle AEB + \angle AZC = 180^\circ. \tag{5} \] From the supposition \( EZ = 2 \cdot ED \), we get that the right triangle \( ZEH \) has \( \angle EZD = 30^\circ \) and \( \angle ZED = 60^\circ \). Thus, we have \( \angle ZEH = 120^\circ \). However, since we have proved that \( \angle BEH = \angle AEZ \), we get that \[ \angle AEB = \angle AEZ + \angle ZEB = \angle ZEB + \angle BEH = \angle ZEH = 120^\circ. \tag{6} \] From (5) and (6), we obtain that \( \angle AZC = 60^\circ \), and thus \[ \angle AEB = 2 \cdot \angle AZC. \tag{7} \]