Given a square $ABCD$, point $E$ is the midpoint of $AD$. Let $F$ be the foot of the perpendicular drawn from point $B$ on $EC$. Point $K$ on $AB$ is such that $\angle DFK = 90^o$. The point $N$ on the $CE$ is such that $\angle NKB = 90^o$. Prove that the point $N$ lies on the segment $BD$. (Matvii Kurskyi)
Problem
Source: 2023 Yasinsky Geometry Olympiad X-XI basic p6 , Ukraine
Tags: geometry, collinear, square