BMO Short List

Let $L^{\prime }$ be the midpoint of $AH$. Then we claim $L^{\prime }$ lies on $\odot ABC$. Indeed, let $D$ be the foot of the $A$-altitude on $BC$. Then: $$AG=GH\Rightarrow \angle G{L}^{\prime }A=90^{\circ }\Rightarrow G{L}^{\prime }\parallel BC\Rightarrow D{L}^{\prime }=\frac{A{L}^{\prime }}{2}=\frac{H{L}^{\prime }}{2}\Rightarrow D{L}^{\prime }=HD$$ where in the last step we have used that if $M$ is the midpoint of $BC$, $AG:GM=2:1$ and that $\angle B$ is obtuse so $H$, $A$ lie on opposite sides of line $BC$. This means that $L^{\prime }$ is the reflection of $H$ in $BC$, which is well-known to lie on $\odot ABC$. Also $AG=GH\Rightarrow \angle G{L}^{\prime }A=90^{\circ }$ so $L^{\prime }$ lies on $\omega$ and hence in fact $L\equiv L^{\prime }$. Let $O$ be the midpoint of $AG$; then $OL\perp LK$. Homothety of factor 2 at $A$ takes $OL\to HG$ so $HG\parallel OL$ and hence $LK\perp HG$. But the centre of $\odot ABC$ lies on $HG$ so this means $K$ is the reflection of $L$ across line $HG$ and hence as $\angle HLG=90^{\circ }$ it follows $\angle HKG=90^{\circ }$.