As $D B=B C=C E$ we have $B I \perp C D$ and $C I \perp B E$. Hence $I$ is orthocenter of triangle $B F C$. Let $K$ be the point of intersection of the lines $B I$ and $C D$, and let $L$ be the point of intersection of the lines $C I$ and $B E$. Then we have the power relation $I B \cdot I K=I C \cdot I L$. Let $U$ and $V$ be the feet of the perpendiculars from $D$ to $E F$ and $E$ to $D F$, respectively. Now we have the power relation $D H \cdot H U=E H \cdot H V$.
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Let $\omega_{1}$ and $\omega_{2}$ be the circles with diameters $B D$ and $C E$, respectively. From the power relations above we conclude that $I H$ is the radical axis of the circles $\omega_{1}$ and $\omega_{2}$.
Let $O_{1}$ and $O_{2}$ be centers of $\omega_{1}$ and $\omega_{2}$, respectively. Then $M B=M C, B O_{1}=C O_{2}$ and $\angle M B O_{1}=\angle M C O_{2}$, and the triangles $M B O_{1}$ and $M C O_{2}$ are congruent. Hence $M O_{1}=M O_{2}$. Since radii of $\omega_{1}$ and $\omega_{2}$ are equal, this implies that $M$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$ and $M, I, H$ are collinear.