Let $ O_1,O_2 $ be the centers of $ \omega_1, \omega_2 $. By Sawayama-Thebault, $ O_1,I,O_2 $ are collinear. Also note that $ IM \parallel AD $. Suppose $ AT $ intersects $ BC $ at $ K $ and the parallel to $ ID $ from $ T $ intersects $ BC $ at $ N $. We'll show that $ (D,M,K,N)=-1 $. We have \[ \frac{DM}{MK}=\frac{AI}{IK} \] and \[ \frac{DN}{NK}=\frac{TI}{TK}=\frac{TA}{TI}=\frac{TA}{TB}=\frac{AC}{CK}=\frac{AI}{IK} \] Hence, $ (D,M,K,N)=-1 $ and since $ TN \parallel DI $, it follows that $ TM $ passes through the midpoint of $ ID $.

Note that $M$ lies on radical axis of the circles. Let $w_1$ touch $AD$ at $X'$, and define $Y'$ similarly, $w_1$ be tangent to $(ABC)$ at $X''$, and define $Y''$ similarly. $X''X \cap Y''Y = T$ and $TX \cdot TX'' = TC^2=TI^2 = TB^2 = TY \cdot TY'' \implies T$ is also on radical axis $\implies TM$ is radical axis of $w_1, w_2$. Next, we note $AX'IX''$ is cyclic by the similar, so $\angle X''X'X = \angle X''AI = \angle X''X'I$ so $I \in XX'$. Hence, $I = XX' \cap YY'$. Let midpoints of $XX', YY'$ be $M_1, M_2$ and midpoint of $X''Y''$ be $M_3$. Note $MM_1M_2M_3$ is a parallelogram, so $M_1M_2$ is bisected by $MM_3$, which is the radical axis of the circles. But, $IM_1DM_2$ is a rectangle, so $MM_3\equiv TM$ bisects $ID$.

Let $P \equiv \omega_1 \cap \odot(\triangle ABC),$ $P' \equiv \omega_2 \cap \odot(\triangle ABC),$ $Q \equiv \omega_1 \cap AD,$ $Q' \equiv \omega_2 \cap AD,$ $T'\equiv PX \cap \odot(\triangle ABC)$ and $I'\equiv QX \cap AI$. Suppose that $K \in BC$ such that $KP$ is tangent to $\omega_1$ and circumcircle of $\triangle ABC$. We have $\angle KXP=\angle KPX=\angle KPT'$ $\implies$ $T'$ is the midpoint of arc $BC$ $\implies$ $T'\equiv T$ $\implies$ $P,X,T$ are collinear $\implies$ $\angle PQX=\angle PXB=\angle PAI'$ $\implies$ $AI'QP$ is cyclic $\implies$ $\angle XI'T=\angle XQD+\angle QAI'=\angle QPX+\angle QPI'=\angle TPI'$ $\implies$ $TI'^2=TX\cdot TP$. Since $T$ is the midpoint of arc $BC$ we deduce that $\angle XBT=\angle TPB$ $\implies$ $TB^2=TX \cdot TP$ $\implies$ $TI'=TB$ $\implies$ $I' \equiv I$ $\implies$ $X,Q,I$ are collinear and $Y,Q',I$ are collinear similarly. We have $\angle XIY=\angle DQ'Y+\angle DQX=180-\frac{1}{2}(\angle ADC+\angle ADB)=90$ $\implies$ $\angle QIM=\angle XIM=\angle QXD=\angle XQD$ $\implies$ $IM\parallel AD$. Now we have $\frac{\sin \angle DIM}{\sin \angle MIT}=\frac{\sin \angle ADI}{\sin \angle DAI}=\frac{AI}{DI}$ $\frac{\sin \angle IDM}{\sin \angle MDT}=\frac{\frac{IC\cdot \sin \frac{1}{2}\angle C}{ID}}{\frac{IT \cdot \sin \frac{1}{2}\angle A}{DT}}=\frac{DT \cdot IC\cdot \sin \frac{1}{2}\angle C}{ID \cdot IT \cdot \sin \frac{1}{2}\angle A}$ So by ceva's sinus form in $\triangle DIT$ and point $M$ we get $\frac{\sin \angle MTI}{\sin \angle MTD}=\frac{DT}{TI}$ hence $TM$ biscets $ID$.

IDMasterz wrote: Note that $M$ lies on radical axis of the circles. Let $w_1$ touch $AD$ at $X'$, and define $Y'$ similarly, $w_1$ be tangent to $(ABC)$ at $X''$, and define $Y''$ similarly. $X''X \cap Y''Y = T$ and $TX \cdot TX'' = TC^2=TI^2 = TB^2 = TY \cdot TY'' \implies T$ is also on radical axis $\implies TM$ is radical axis of $w_1, w_2$. Next, we note $AX'IX''$ is cyclic by the similar, so $\angle X''X'X = \angle X''AI = \angle X''X'I$ so $I \in XX'$. Hence, $I = XX' \cap YY'$. Let midpoints of $XX', YY'$ be $M_1, M_2$ and midpoint of $X''Y''$ be $M_3$. Note $MM_1M_2M_3$ is a parallelogram, so $M_1M_2$ is bisected by $MM_3$, which is the radical axis of the circles. But, $IM_1DM_2$ is a rectangle, so $MM_3\equiv TM$ bisects $ID$. There's a typo, $M_3$ should be the midpoint of $X'Y'$.

IDMasterz wrote: Note that $M$ lies on radical axis of the circles. Let $w_1$ touch $AD$ at $X'$, and define $Y'$ similarly, $w_1$ be tangent to $(ABC)$ at $X''$, and define $Y''$ similarly. $X''X \cap Y''Y = T$ and $TX \cdot TX'' = TC^2=TI^2 = TB^2 = TY \cdot TY'' \implies T$ is also on radical axis $\implies TM$ is radical axis of $w_1, w_2$. Next, we note $AX'IX''$ is cyclic by the similar, so $\angle X''X'X = \angle X''AI = \angle X''X'I$ so $I \in XX'$. Hence, $I = XX' \cap YY'$. Let midpoints of $XX', YY'$ be $M_1, M_2$ and midpoint of $X''Y''$ be $M_3$. Note $MM_1M_2M_3$ is a parallelogram, so $M_1M_2$ is bisected by $MM_3$, which is the radical axis of the circles. But, $IM_1DM_2$ is a rectangle, so $MM_3\equiv TM$ bisects $ID$. Could you please tell me that why is AX'IX'' concyclic?

Now I have a new solution for this. By Kanazawa mountain theorem we conclude that I lies on XX' and also YY'. Denote the intersection of TM and X'Y' as K, then IKMD is a parallelogram (by simple angle chasing). Thus we get TM goes through the midpoint of ID, as desired.

Here is an alternate solution:By Arichmetic lemma it is clear that $TM$ is thr radical axis of $\omega_1 ,\omega_2$.So it is enough to say midpoint of $ID$ also lies on the radical axis.This is clear by considering radical center of circles $\omega_1 ,\omega_2$ and the point $D$ and swayama lemma.

lemma: $\triangle ABC $ with $BA=BC$ $D \in AB , F \in BC, E = FD \cap AC$ such that $DB=BF$ then the midpoints of $AD, CF, EB$ are collinear now let's return to our problem: let $\omega_1$,$\omega_2$ touch $AD$ in $Z,W$ it's well-known that $I=ZX \cap YW $ and we have $POW_T(\omega_1)=POW_T(\omega_2)$ and from the lemma midpoint of $ID$ is on the radical axis of two circle and so are $M,T$