My solution: Let $AI\cap BC=X$. By Csarnot's theorem on $\triangle IMX$ perpendiculars from $M$ to $IX$,$D$ to $IM$ and $A$ to $MX$ are concurrent iff $AX^2-AM^2+DM^2-DI^2+MI^2-MX^2=0$ which is straightforward calculation.

Synthetic: As we see many cyclic quadrilaterals, we imediatelly think of radical axis... Letting $E $ be the tangency point of $A $-excircle, and $S $ the foot of altitude from $D $ to $AE $, we easily find that $IM $ bissects $DS $. So, $\angle ISM =\angle IDM=90$, so if $L$ is the foot of altitude from $M$ to $AI $, $IMDSL $ is cyclic, and also if $H $ is foot of perpendicular from $A $ to $BC $, quadrilaterals $AHDS $ and $AHLM $ are cyclic from easy angle chase. So their radical axes are concurrent, so $AH, ML$, and $DS $ are concurrent as required.

My solution: Let $H$ is the foot of perpencicular of $A$ in $BC$ and $G$ is the foot of perpendicular of $M$ en $AI$ and let $X=AH\cap MG$ Let $D'$ the reflection of $D$ in $I$ and $AD'$ cuts $BC$ in $Y$ $\Longrightarrow$ $IM\parallel AY$. Let $Z=ID\cap MG$ $\Longrightarrow$ $U$ is the ortocenter of $\triangle IMZ$ Since $AI\parallel AH$ and $IM\parallel AY$ we get $\tfrac{HD}{DU}=\tfrac{AI}{IU}=\tfrac{MY}{UM}$, but $MY=MD$ $\Longrightarrow$ $\tfrac{HD}{DU}=\tfrac{MD}{MU}$ $\Longrightarrow$ $\tfrac{HD}{DM}=\tfrac{DU}{UM}...(\star)$ Since $XH\parallel ZD$ $\Longrightarrow$ $\tfrac{HD}{DM}=\tfrac{XZ}{ZM}...(\star\star)$ $\Longrightarrow$ by $(\star)$ and $(\star)(\star)$ we get $\tfrac{DU}{UM}=\tfrac{XZ}{ZM}$ $\Longrightarrow$ $XD\parallel UZ$. Since $U$ is the ortocenter of $\triangle IMZ$ $\Longrightarrow$ $UZ\perp IM$ $\Longrightarrow$ $XD\perp IM$ hence the perpendiculars from $M, D, A $ to $AI, IM, BC $ respectively are concurrent in $X$

nice and ez

Attachments:

Dear Mathlinkers, 1. (1) the circle with diameter IM 2. Reim's theorem with (1) and incircle give a parallelism to AH 3. Pascal's theorem in a spacial case leads to the result... Sicerely Jean-Louis

Here is a very short and elegant solution found with Neothehero. Construct $K$ so that $AIDK$ is parallelogram. By homothety at $D$, line $IM$ bisects $AD$ hence it goes through $K$. Observe that the concurrency point is orthocenter of $\triangle KDM$ so we are done.