10. Consider first a triangle. It can be decomposed into $k=3$ cyclic quadrilaterals by perpendiculars from some interior point of it to the sides; also, it can be decomposed into a cyclic quadrilateral and a triangle, and it follows by induction that this decomposition is possible for every $k$. Since every triangle can be cut into two triangles, the required decomposition is possible for each $n \geq 6$. It remains to treat the cases $n=4$ and $n=5$. $n=4$. If the center $O$ of the circumcircle is inside a cyclic quadrilateral $A B C D$, then the required decomposition is effected by perpendiculars from $O$ to the four sides. Otherwise, let $C$ and $D$ be the vertices of the obtuse angles of the quadrilateral. Draw the perpendiculars at $C$ and $D$ to the lines $B C$ and $A D$ respectively, and choose points $P$ and $Q$ on them such that $P Q \| A B$. Then the required decomposition is effected by $C P, P Q, Q D$ and the perpendiculars from $P$ and $Q$ to $A B$. $n=5$. If $A B C D$ is an isosceles trapezoid with $A B \| C D$ and $A D=B C$, then it is trivially decomposed by lines parallel to $A B$. Otherwise, $A B C D$ can be decomposed into a cyclic quadrilateral and a trapezoid; this trapezoid can be cut into an isosceles trapezoid and a triangle, which can further be cut into three cyclic quadrilaterals and an isosceles trapezoid. Remark. It can be shown that the assertion is not true for $n=2$ and $n=3$.