JBMO-SL 2005 G4

Let $\mathrm{ABC}$ be an isosceles triangle such that $A B=A C$ and $\angle \frac{A}{2}<\angle B$. On the extension of the altitude $\mathrm{AM}$ we get the points $\mathrm{D}$ and $\mathrm{Z}$ such that $\angle C B D=\angle A$ and $\angle Z B A=90^{\circ}$. $\mathrm{E}$ is the foot of the perpendicular from $\mathrm{M}$ to the altitude $\mathrm{BF}$ and $\mathrm{K}$ is the foot of the perpendicular from $\mathrm{Z}$ to $\mathrm{AE}$. Prove that $\angle K D Z=\angle K B D=\angle K Z B$.