Question 38
Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle \frac{A}{2} \lt \angle B$. On the extension of the altitude $AM$ we get the points $D$ and $Z$ such that $\angle CBD = \angle A$ and $\angle ZBA = 90^{\circ}$. $E$ is the foot of the perpendicular from $M$ to the altitude $BF$ and $K$ is the foot of the perpendicular from $Z$ to $AE$. Prove that $\angle KDZ = \angle KBD = \angle KZB$.