4. Given a square $A B C D$. Point $N$ lies on side $A D$, such that $A N$ : $N D=2: 3$, point $F$ lies on side $C D$ and $D F: F C=1: 4$, point $K$ lies on side $A B$, such that $A K: K B=1: 4$. Find the angle $K N F$.

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Solution. Mark point $E$ on side $B C$ such that $B E: E C=2: 3$. Let $M-$ be the midpoint of $K E$. Then $\triangle K M N-$ is an isosceles right triangle and $\triangle M N F-$ is an isosceles triangle. Therefore, $\angle K N F=$ $\angle K N M+\angle M N F=45^{\circ}+90^{\circ}=135^{\circ}$.