Solved with @primesarespecial and @theproblemissolved as far as I can remember.

Sketch/ Key claim/ wtv

Let $DP$ be the tangent from $D$ to $(ABC)$. Obviously $EF$ is the perpendicular bisector of $AD$, so $PD$ is tangent to $(AMD)$ iff $MD$ bisects $\angle APD$. But $\triangle APD$ is isosceles, so we need $MD \perp AP$, so proving $\angle FMD = \angle PAD$ is sufficient. We have that $\angle AMF= \angle FMD$, so we need $\angle PAD= \angle AMF$. Add the midpoint $N$ of $BC$; notice that by angle chasing $\triangle AEF \cup \{M\} \sim \triangle ABC \cup \{N\}$, so $\angle AMF= \angle ANC= \angle ABP =\angle PAD$, done (we used that $AN$ and $AP$ are isogonal in $\angle BAC$).

Let $O$ be the circumcenter of $(ABC)$ and $DG$ be the other tangent line through $D$ to $(ABC)$. Notice that $\triangle ADE\sim\triangle AOC$, so $DE\perp AC$. And $OF\perp AC$ for the obvious reason. Similarly $DF\perp AB\perp OE$. So $EDFO$ is a parallelogram. So $M$ be the midpoint of $DO$ which bisects $\angle ADG$. Hence $\angle DAM=\angle ADM=\angle MDG$. Done!

Let $\omega = \odot(ABC)$ and $O$ be its center; $K$ be the point of $\omega$ distinct from $A$ such that $DK$ is tangent to $\omega$; $N$ be midpoint of $BC$; $S$ be reflection of $A$ in line $BC$ and $T$ be reflection of $A$ in perpendicular bisector of segment $BC$ (or line $ON$). We want to show $DK$ is also tangent to $\odot(AMD)$. Lemma 1: (Known) Let $ABC$ be a triangle and $D$ be any point on side $BC$. Let $E,F$ be circumcenters of $\triangle ABD, \triangle ACD$, respectively. Then, $$ \triangle AEF \stackrel{+}{\sim} \triangle ABC $$Proof: Note $EA = EB$, $FA = FC$. Further, $$ \measuredangle BAE = 90^\circ - \measuredangle ADB = 90^\circ - \measuredangle ADC = \measuredangle CAF $$It follows $$ \triangle AEB \stackrel{+}{\sim} \triangle AFC $$which implies $\triangle AEF \stackrel{+}{\sim} \triangle ABC$, as desired. $\square$ [asy][asy] size(250); pair B=dir(-165),C=dir(-15),A=dir(120),O=(0,0),N=1/2*(B+C),S=2*foot(A,B,C)-A,K=IP(N--S,unitcircle),T=2*foot(A,N,O)-A,D=2*A*K/(A+K); pair E=circumcenter(A,B,D),F=circumcenter(A,D,C),M=1/2*(E+F); fill(A--D--M--A--cycle,cyan+white+white); fill(A--S--N--cycle,green+white+white); draw(A--D--K,brown); draw(E--F,magenta); draw(D--C,red); draw(S--T); draw(A--M--D,purple); draw(S--A--N,purple); draw(A--B--S--C--A,orange+linewidth(0.8)); draw(A--E--D--F--A,fuchsia+linewidth(0.8)); draw(unitcircle,royalblue); draw(circumcircle(A,O,K),grey); dot("$A$",A,dir(90)); dot("$B$",B,dir(-110)); dot("$C$",C,dir(C)); dot("$O$",O,dir(60)); dot("$N$",N,dir(-60)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$K$",K,dir(-70)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(-140)); [/asy][/asy] Let $f$ be the spiral similarity centered at $A$ taking $E \to B$ and $F \to C$. Note $f(M) = N$. Further, as $D$ is reflection of $A$ in line $EF$, so $f(D) = S$. It follows $$ AEFMD \stackrel{+}{\sim} ABCNS $$In particular, this gives $$ \angle AMD = \angle ANS \qquad \qquad (1)$$Claim 2: Points $S,K,N,T$ are collinear. Proof: As $AK$ is symmedian (in $\triangle ABC$), so it follows $K \in NT$. Also, as $N$ is circumcenter of $\triangle TAS$ with $\angle TAS = 90^\circ$, it follows $N \in TS$. It follows both $K,S$ lie on $TN$, as desired. $\square$ Now note the points $A,O,N,K,D$ lie on circle with diameter $DO$. We obtain $$ \angle AMD = \angle ANS = \angle ANK = 180^\circ - \angle ADK $$It follows $DK$ is tangent to $\odot(AMD)$ at $D$, which completes the proof. $\blacksquare$

Clearly $EF$ is the perpendicular bisector of $AD$ and as $M$ lies on $EF$, then $MA = MD$. Let $O$ be the circumcenter of triangle $ABC$. Claim: $\angle EOF = \angle BAC$ Proof: It's easy see that $EO$ is perpendicular bisector of $AB$ and $FO$ is perpendicular bisector of $AC$. Then if $I = EO \cap AB$ and $J = FO \cap AC$ we have that $AIOJ$ is cyclic, where $\angle BAC = \angle IAJ = 180^{\circ} - \angle IOJ = \angle EOF \square$ Claim: $DEOF$ is a parallelogram Proof: It's suffices show that $\angle EDF = \angle EOF$ and $\angle DEO = \angle DFO$. Notice that $\angle DEO = \angle DEB + \angle BEO = 2 \angle DAB + \frac{\angle BEA}{2} = 2\angle DAB + \angle ADB$. Now, $\angle DFO = \angle DFA + \angle AFO = 2\angle DCA + \frac{AFC}{2}$ but $AD$ is tangent to $\odot (ABC)$ at $A$, then $\angle DAB = \angle DCA$. Therefore $\angle DFO = 2\angle DCA + \frac{AFC}{2} = 2\angle DAB + \angle ADC = 2\angle DAB + \angle ADB = \angle DEO$. We observe that $\angle EDF = \angle EDB + \angle BDF = \angle EDB + \angle CDF = (90^{\circ} - \angle DAB) + (90^{\circ} - (180^{\circ}- \angle DAC)) = \angle DAC - \angle DAB = \angle BAC = \angle EOF$ $\square$ Thus $DEOF$ is a parallelogram whose diagonals intersect at their midpoint, which is $M$. Then $D$, $M$ and $O$ are collinear. Let $T$ be the tangency point from $D$ to the $\odot (ABC)$ different of $A$; clearly $DO$ is perpendicular bisector of $AT$, then $\angle TDO = \angle ODA = \angle MDA = \angle MAD$ but $\angle TDM = \angle TDO = \angle MAD$ thus $TD$ is tangent to the circumcircle of $AMD$ and is also tangent to the circumcircle of $ABC$, as desired $\blacksquare$

Attachments:

Just bash it Consider $(ABC)$ as the unit circle and let the other tangent from $D$ to $(ABC)$ touch the circle at $X$ $d=\frac{2ax}{a+x} \implies x=\frac{ad}{2a-d}$ $$D=AA\cap BC \implies d=\frac{a(ab+ac-2bc)}{a^2-bc}$$Since $E$ is the circumcenter of $(ABD) : $ $$e=\begin{vmatrix} a & a\overline{a} & 1\\ b & b\overline{b} & 1\\ d & d\overline{d} & 1\\ \end{vmatrix}\div \begin{vmatrix} a & \overline{a} & 1\\ b & \overline{b} & 1\\ d & \overline{d} & 1\\ \end{vmatrix}=\begin{vmatrix} a-b & 0 & 0\\ b & 1 & 1\\ d & d\overline{d} & 1\\ \end{vmatrix}\div \begin{vmatrix} a-b & \frac{a-b}{-ab} & 0\\ b & \frac{1}{b} & 1\\ d & \overline{d} & 1\\ \end{vmatrix}$$$$e=\frac{ab(d\overline{d}-1)}{ab\overline{d}+d-a-b}=\frac{ab(2a^3b+2a^3c-4a^2bc+2abc^2+2ab^2c-a^2b^2-b^2c^2-a^2c^2-a^4)}{(a^2-bc)(2a^2b^2-ab^2-2abc+a^2c+b^2c-a^3)}$$$$e=\frac{ab(a-b)^2(a-c)^2}{(a^2-bc)(a-b)^2(a-c)}=\frac{ab(a-c)}{a^2-bc}$$Similarly $: f=\frac{ac(a-b)}{a^2-bc}$ $$\implies m=\frac{e+f}{2}=\frac{a(ab+ac-2bc)}{2(a^2-bc)}=\frac{d}{2}$$Now we wanna prove that $DX$ is tangent to $(AMD)$ wich equivalent to : $$\angle ADX +\angle DMA =180^\circ\iff \Delta :=\frac{a-d}{x-d}\times\frac{d-m}{a-m}\in\mathbb{R}$$$$\iff \Delta =\frac{(a-d)(d-\frac{d}{2})}{(\frac{ad}{2a-d}-d)(a-\frac{d}{2})}=\frac{d(a-d)}{d(d-a)}=-1\in\mathbb{R}$$$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ \blacksquare$

Let $O$ be the circumcenter of $(ABC)$. Note that $EF$ is the perpendicular bisector of segment $AD$. Also, lines $EO$ and $FO$ are the perpendicular bisectors of $AB$ and $AC$ respectively. Notice that, $$\angle DEA = 360 - 2 \angle DBA = 2 \angle B$$ and, $$\angle DFC = 360 - 2\angle DAC = 360 - 2 \angle DBA = 2\angle B.$$ So, $\triangle DEA \sim \triangle DFC$. Also, $$\angle DEB = 2 \angle DAB = 2 \angle C$$ and, $$\angle DFA = 2 \angle DCA = 2 \angle C.$$ So, $\triangle DEB \sim \triangle DFA$ too. Then, we make the following key claim. Claim : $\triangle ABC \sim \triangle EFA \sim \triangle DEF \sim \triangle EOF.$ Proof : Let $K$ be the midpoint of $AD.$ Then, $$\angle FEA = \frac{\angle DEA}{2} = \angle B.$$ And, $$\angle EFA = 90 - \angle DAF = 90 - \angle EBD = \frac{\angle DEB}{2} = \angle C.$$ So, $\triangle ABC \sim \triangle EFA$. Note that $\triangle EFA \cong \triangle DEF$ so, $\triangle ABC \sim \triangle EFA \sim \triangle DEF$. Now, let $P$ and $Q$ be the midpoints of $AB$ and $AC$ respectively. Then, it is clear that $E, P, O$ and $F, Q, O$ are collinear. Thus, $$\angle EOF = \angle POF = 180 - \angle POQ = \angle A.$$ And, $$\angle FEO = \angle FEA - \angle AEO = \frac{\angle AED}{2} - \angle AEP = \angle B - \frac{\angle AEB}{2} = \angle B - \angle ADB = \angle DAB = \angle C.$$ So, $\triangle ABC \sim \triangle FOE$ as well. Establishing the claim. Next, we will show that $DEOF$ is a parallelogram. This follows by similarities i.e. $\angle OEF = \angle EFD$ and $\angle DEF = \angle EFO$. Which implies $EO \parallel DF$ and $DE \parallel OF$ implying $DEOF$ parallelogram. This means, as $M$ is the midpoint of $EF$, it must also be the midpoint of $DO$. So, $MA = MD = MO$. Now, let $T$ be the tangency point such that $TD$ is tangent to $(ABC).$ So, $DO$ is the perpendicular bisector of $AT$ which gives $MA = MT$ too. So points $A, O, T, D$ are concylic with $M$ being center of that circle. Now, $$\angle DMT = 2 \angle DAT = 2 \angle DTA = \angle DMA.$$ So by $S.A.S,$ $\triangle DMT \cong \triangle DMA$ which implies $\angle TDM = \angle MAD$ which gives our desired tangency.

Very nice problem Claim 1:$ \triangle AEF \stackrel{+}{\sim} \triangle ABC $ Proof: By Salmon's Theorem with $\triangle ABC$ and $D$, $ \triangle AEF \stackrel{+}{\sim} \triangle ABC $ $\square$ By spiral similarity with center A. $ AEFMD \stackrel{+}{\sim} ABCZX $. So $ \angle AMD = \angle AZX $ Let: $O: \text{circumcenter of } \triangle ABC$ $X: \text{ the reflexion of } A \text{ in } BC$ $Y: Y\in \odot ABC \text{ and } DY \text{ tangent to } \odot ABC$ $Z: \text{ the midpoint of } BC$ $R: \text{ the reflexion of } A \text{ in bisector of } BC$ Claim 2: $X,Y,Z,R$ are collinear. Proof: $ABYC$ armonic quadrilateral, so $\angle AZB=\angle BZY$, but $\angle AZB=\angle BZX$, so $X,Y,Z$ are collinear. But $\triangle ABZ \cong \triangle RCZ\Rightarrow \angle RZC=\angle AZB=\angle BZY$, so $Y,Z,R$ are collinear. So $X,Y,Z,R$ are collinear. $\square$ Claim 3: $D,A,O,Z,Y$ are concyclic. Proof: It's easy to see $DA\perp AO,OZ\perp DZ\text{ and } OY\perp DY$, So $D,A,O,Z,Y$ are concyclic with diameter $DO$. $\square$ Finishing: $ \angle AMD = \angle AZX =2\alpha$, but $OA= OY\Rightarrow \angle AOD=\angle DOY=\alpha$ so $\angle ODY= 90^{\circ}-\angle DOY=90^{\circ}-\alpha$ So: $\angle DMA=90^{\circ}-\alpha$, so $DY $ is tangent to $\odot AMD$, As desired.

Let $K$ be the second point such that $DK$ is tangent to $(ABC)$. Now, we have the following key claim. Claim : Quadrilateral $DEOF$ is a parallelogram. Proof : The angle chasing is a bit messy but we can grind through. We use the well known fact that the radical axis is perpendicular to the line joining the centers to conclude that $EO\perp AB$,$FO\perp AC$ and $EF\perp AD$. This will be used throughout this proof. Now, $\measuredangle FOE = \measuredangle CAB$. Then, \[2\measuredangle EDB = \measuredangle DEB = 2\measuredangle DAB = 2\measuredangle ACB = 90+ \measuredangle ACB\]Thus, $\measuredangle EDB= 90 + \measuredangle ACB$. Also, \[2\measuredangle BDF = \measuredangle CFD = 2\measuredangle CAD - 180\]Then, we have that $\measuredangle BDF +90 = \measuredangle CAD$. Thus, \[\measuredangle EDF = \measuredangle EDB + \measuredangle BDF = 90 + \measuredangle ACB + \measuredangle CAD - 90 = \measuredangle CAB\]Thus, \[\measuredangle FOE = \measuredangle EDF\]Further note that, \[\measuredangle OED = 360 - ( \measuredangle BDE +90 + \measuredangle ABD) = 180 + \measuredangle BCA + CBA = 2\measuredangle CBA + \measuredangle BAC\]with a similar angle chase, we obtain that $\measuredangle DFO = \measuredangle BAC + 2\measuredangle CBA$. Thus, \[\measuredangle OEF = \measuredangle DFO\]as well. This implies the required result. Claim : $DKOA$ is a cyclic quadrilateral. Proof : Simply note that $\measuredangle OAD = 90^\circ = \measuredangle OKD$. Thus, $DKOA$ is indeed cyclic. Now, since parallelograms have diagonals which bisect each other, $M$ is also the midpoint of $DO$. But, the midpoint of $DO$ is the center of $(AOKD)$! Thus, $ MD =MA=MK$. So, \[\measuredangle MDK = \measuredangle ADK + \measuredangle KDM = \measuredangle ADM = \measuredangle MAD\]This means that the line tangent to $(ABC)$ through $D$ which does not pass through $A$, is in fact tangent to $(AMD)$. Since there exists exactly one tangent to a circle from a point on the circle, we can conclude that indeed, the line tangent to the circumcircle of $AMD$ through $D$ is also tangent to the circumcircle of $ABC$ and we are done.

Solved in 5 minutes (:skull:), nice problem though Let $N$ the midpoint of $BC$ and $K$ a point in $(ABC)$ such that $AK$ is symedian, then since $K \to N$ in $\sqrt{bc}$ invert we have that from easy angle chase that $ADKN$ is cyclic, redefine $M$ as the center of this cyclic, then we will prove that $M$ is midpoint of $EF$ which clearly finishes as then trivially $\angle MDN=\angle ADM=\angle DAM$ which implies the tangency. Now draw perpendicular bisectors of $DB,DC,DN$, notice that the distance from $DB$ to $DN$ is the same as $DN$ to $DC$ by the midpoint, and by radax we have $E,F,M$ colinear so by thales we have $EM=MF$ as desired thus we are done .

Why has nobody trigbashed this? Let $K$ be the intersection with the $A$-symmedian and the circumcircle of $(ABC)$. $K$ lies on the $A$-Apollonian circle, hence $DK$ is tangent to $(ABC)$. We claim that it is also tangent to $(AMD)$ Claim. $\angle EAM=\angle CBK$ and $\angle FAM=\angle BCK$ Proof: these clearly have the same sums so it suffices to prove that $$\frac{\sin(\angle EAM)}{\sin(\angle FAM)}=\frac{\sin(\angle CBK)}{\sin(\angle BCK)}$$ From ratio lemma we have $$\frac{\sin(\angle EAM)}{\sin(\angle FAM)}=\frac{FA}{EA}=\frac{AD\sin(\angle ACB)}{AD\sin(\angle ABC)}=\frac{AB}{AC}$$ The second ratio also follows from ratio lemma $$\frac{\sin(\angle CBK)}{\sin(\angle BCK)}=\frac{\sin(\angle CAK)}{\sin(\angle BAK)}=\frac{AB}{AC}$$ Hence the claim is proven. Now easy angle chase yields $\angle DAM=\angle ADK/2$. $\blacksquare$

somehow this is rated d6 in modsmo but i think this has to be atleast a d7. anyways here’s my sol: Let $T$ be foot of $A$-perpendicular on $BC$ and $N$ the midpoint of $BC$. Let the tangent from $D$ to $(ABC)$ touch it in $K$ different from $A$. CLAIM 1: $\triangle ABC \sim \triangle AEF$ PROOF: Indeed, observe that $\angle EAF=\angle EAD +\angle DAF=90-\angle B +90-\angle C=\angle A$. Also observe that $\angle EBA=\angle 90-\angle ADB=\angle AFC$. Using $EA=EB$ and $FA=FC$, we get $\triangle EAB \sim \triangle FAC \implies \frac{AE}{AF}=\frac{AB}{AC} \implies \triangle AEF \sim \triangle ABC$ CLAIM 2: $\angle DM =\angle TAN$ PROOF: This is true since under a spiral similarity at $A$ mapping $\triangle AEF$ to $\triangle ABC$, $M \mapsto N$ and $AD \cap EM \mapsto T$ [using EM $\perp AD$] CLAIM 3: $AK$ is the symmedian of $\triangle ABC$ PROOF: This is true since $(A,K;B,C)=-1$, a well known fact Finally, we shall finish by chasing angles: $\angle MDK=180- \angle MAD -\angle DAK- \angle- \angle DKA$ $=180- \angle TAN-2 \angle DKA$ =$180- \angle TAN - 2( \angle C + \angle BAK)$ $= 180- \angle TAN- 2( \angle C + \angle NAC)$ $=180- \angle TAN- 2( \angle C + 90- \angle C - \angle TAN)$ $180- \angle TAN- 2( 90-\angle TAN)$ $= \angle TAN=\angle MAD$ and we are done. $\blacksquare$