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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.

(Ukraine)

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