Let $X$ be the midpoint of $BC$. It is well-known that $D=XH\cap (ABC)$. Let $H'$ be the reflection of $H$ over $X$. It is well-known that $H' \in (ABC)$ (easy angle chase). Also, since $(BHC)$ is simply the reflection of $(ABC)$ over $BC$, we know $N$ is the reflection of $N'$, the arc midpoint of big arc $BC$ on $(ABC)$, over $BC$. Now, since $H',D,M,N'$ all lie on $(ABC)$, \[XH\cdot XD = XH'\cdot XD = XM\cdot XN'= XM\cdot XN\]and we are done.

Did anyone use $\sqrt{bc}$ inversion

It's easy to show that $N$ is the reflection of the midpoint of arc BAC in$(ABC)$. Thus, $LM\cdot LN=LC\cdot LB=LH\cdot LD$. QED.