Let $ABC$ be an isosceles triangle $(AB=AC)$ so that $\angle A< 2 \angle B$ . Let $D,Z $ points on the extension of height $AM$ so that $\angle CBD = \angle A$ and $\angle ZBA = 90^\circ$. Let $E$ the orthogonal projection of $M$ on height $BF$, and let $K$ the orthogonal projection of $Z$ on $AE$. Prove that $ \angle KDZ = \angle KDB = \angle KZB$.