Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order. Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other. Proposed by Ukraine