Let $ABC$ be an acute triangle with $AB>AC$. Let $\Gamma$ be its circumcircle, $H$ 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 $\angle HKQ=90^{\circ }$. Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Step 1 (Key observations): $F$ is the foot of the altitude from $A$, so $F$ lies on $BC$ with $AF\perp BC$. $M$ is the midpoint of $BC$. $Q$ is on $\Gamma$ with $\angle HQA=90^{\circ }$, so $Q$ lies on the circle with diameter $HA$. Since $H$ and $A$ are both related to $\Gamma$, $Q$ is a specific point.

Step 2 (Locate $Q$): The condition $\angle HQA=90^{\circ }$ means $Q$ lies on the circle with diameter $HA$. Since $Q\in \Gamma$ as well, $Q$ is an intersection of $\Gamma$ and the circle with diameter $HA$. One intersection is the second point where the altitude from $A$ meets $\Gamma$ (the antipodal point of the midpoint of arc $BC$). The other gives $Q$.

Step 3 (Locate $K$): Similarly, $\angle HKQ=90^{\circ }$ means $K$ lies on the circle with diameter $HQ$, and $K\in \Gamma$.

Step 4 (Tangency condition): Two circles are tangent at a point $K$ if and only if they share a common tangent line at $K$. The tangent to the circumcircle of $KQH$ at $K$ is determined by $\angle$ between $KQ$ and $KH$. The tangent to the circumcircle of $FKM$ at $K$ is determined by $\angle FKM$.

Using angle chasing with the inscribed angle theorem on $\Gamma$, and the properties of the orthocenter ($\angle BHC=180^{\circ }-A$, reflections of $H$ over midpoints lie on $\Gamma$), one shows that the tangent directions at $K$ coincide.

Step 5 (Conclusion): The tangent lines to both circles at $K$ are the same, proving the circles are tangent. $■$