Let $F$ be midpoint of $AC$ $\implies$ $MF||BC$. Also, obviously $KHMFC$ is cyclic. Let $K,L$ be foot from $D,E$ to $BC,AB$ and Let $N$ be midpoint of $DE$. By Three Tangents Lemma, $NK$ is Tangent to $\odot (BHK)$. But, $\angle MFH$ $ =$ $\angle MKH$ $=$ $\angle HBK$ $\implies$ $MK$ is also tangent to $\odot (BHK)$ $\implies$ $N$ $\in$ $MK$. Since, $MN||AB$ $\implies$ $K$ is midpoint of $AE$. Thus, $\Delta BDE$ is Equilateral. So, $\angle ABC= 60^{\circ}$ $\qquad \blacksquare$