$\textbf{LEMMA:-}$ $ABC$ be a triangle and let $\Delta M_AM_BM_C$ be the Medial Triangle of $\Delta ABC$. Let the tangent to $\odot(ABC)$ at $C$ intersect $\overline{M_BM_C}$ at $P$ and let $\overline{M_AM_B}$ hit $\odot(ABC)$ at $\{X\}$. Then $\{\overline{XA},\overline{XP}\}$ are Isogonal Conjugates WRT $\Delta CXM_C$. Let $\overline{XP}\cap\overline{BC}=\{Q\}$ and $\overline{AC}\cap\overline{XM_C}=\{R\}$. Then $\frac{RX}{RM_C}=\frac{M_BX}{M_AM_B}=\frac{PX}{PQ}$. Now, $$\frac{PX}{PQ}=\frac{CX}{CQ}\cdot\frac{\sin\angle PCX}{\sin\angle QCP}=\frac{CX}{CQ}\cdot\frac{\sin\angle XAR}{\sin\angle M_CAR}=\frac{RX}{RM_C}=\frac{XA}{AM_C}\cdot\frac{\sin\angle XAR}{\sin\angle M_CAR}\implies\frac{CX}{CQ}=\frac{AX}{AM_C}$$also we know that $\measuredangle X_AMC=\measuredangle XCQ\implies\Delta XAM_C\cup R\stackrel{+}{\sim}\Delta XCQ\cup P$. So, $\{\overline{XA},\overline{XP}\}$ are Isogonal Conjugates WRT $\Delta CXM_C$. ____________________________________________________________________________________________ Coming back to the problem. Let the tangent to $\odot(ABC)$ at $A$ intersect $\odot(ABC)$ at $V$ and let $\overline{M_AM_B}\cap\overline{AV}=\{U\}$. Now by Dual of Desargues Involution Theorem on the Complete Quadrilateral $\mathbf{Q}\equiv\{\overline{M_AM_B},\overline{AM_B},\overline{AV},\overline{M_AV}\}$ we get that there is an Involution $\Phi$ which swaps $\{(\overline{PA},\overline{PM_A}),(\overline{PM_B},\overline{PV}),(\overline{PU},\overline{PC})$. Now by $\textbf{LEMMA}$ we already know that $\{\overline{PU},\overline{PC}\}$ are Isogonal Conjugates WRT $\Delta APM_A$. So, from $\Phi$ $(\overline{PM_B},\overline{PB})$ are Isogonal Conjugates too WRT $\Delta APM_A$. So, $\measuredangle VPM_A=\measuredangle APM_C=\measuredangle PQM_A\implies\overline{PV}$ is tangent to $\odot(PM_AQ)$. $\blacksquare$

Nice problem! Let $M_b M_c$ intersect $\odot (ABC)$ at $R$, and $AR$ intersect $BC$ at $S$. Also, define $K$ as the intersection of the tangent to $\odot (RSM_a)$ at $R$ and the tangent to $\odot (ABC)$ at $A$. Then, $\angle KAP=180^\circ -\angle ARP=180^\circ -\angle ASQ = \angle KRM_a$. $\frac{KR}{KA}=\frac{\sin\angle RAK}{\sin\angle KRA}=\frac{\sin\angle APR}{\sin\angle M_a RP} =\frac{\triangle APR}{\triangle RM_a P}\times\frac{PM_a }{AP}=\frac{RM_a }{AP}$ So, $\triangle KRM_a \sim\triangle KAP \rightarrow \triangle KPM_a \sim \triangle KAR$ Then $\angle KPM_a=\angle KAR=\angle APR=\angle PQM_a $, which shows that $KP$ is a tangent to $\odot (PQM_a)$ Also, $\angle KM_a P=\angle KRA =\angle RM_a C=\angle BM_aP$,and hence, $K$ lies on $BC$

$\textbf{Notation:}$ Let $K$ be the meeting point of $BC$ and the tangent at $A$ to $(ABC)$. $A'\in (ABC)$ such that $AA'\parallel BC$. $N$ the midpoint of $AA'$ $\textbf{Solution:}$ 1- with just angle working you get $\triangle AQK\sim A'AP$, then $\angle KPQ = \angle PNA$ 2- line $PM_cM_a$ bisects $NM_a$ then $\angle PNA = \angle PM_aB$ 3- Combining the previous $\angle KPQ = \angle PM_aB$, then $KP$ is tangent to $(PQM_a)$.

Let $M_bM_c$ meets $(ABC)$ again at $X$. Let $AP\cap XM_a=R$, and $AX\cap PM_a=S$ Since $PX$ passes through the midpoint of $AM_a$, by Ceva on $\triangle APM_a$ and point $X$ we have $AM_a\parallel RS$. By Menelaus on $\triangle RPM_a$ and $\overline{AXS}$, we have $\frac{RA}{AP}\cdot\frac{PS}{SM_a}\cdot\frac{M_aX}{XR}=1$ $\frac{XM_a}{SM_a}=\frac{PM_a}{SM_a}=\frac{PA}{RA}$, so $PS=XR$. Let perpendicular bisector of $PR$ and $XS$ meet at $K$. $PK=RK$ and $XK=SK$, so $\triangle KSP\equiv\triangle KXR$, and therefore $\triangle KSX\sim\triangle KPR$ $\angle KSX=\angle KPR$, $\angle KXS=\angle KRP$ so we have that $APSK, ARKX$ is cyclic. It shows that $K$ is a Miquel point of quadrilateral $APM_aX$. $\angle PM_aX=\angle PKR$, so $\triangle PM_aX\sim\triangle PKR\sim\triangle SKX$ $\angle XM_aC=\angle M_aXP=\angle KXS=\angle XSK=\angle XM_aK$, so $K$ is on line $BC$. $\angle KAX=\angle KPS=\angle APX=\angle AQK$, so $KA, KP$ is the tangent of $(APX), (PQM_a)$, respectively. Therefore, $K$ is the concurrency point. Remark : This is the similar idea to 2019 RMM #2

We let $T$ be the intersection of the tangent through $A$ with $BC$ and wish to show a tangency. Let $K$ be the intersection of the $A$-symmedian with $(ABC)$. Now it is well-known that $KM_A \cap (ABC) = A'$ is such that $AA'\parallel BC$. This implies (after an easy angle chase) that $QPM_AK$ is cyclic. Now we perform $\sqrt{bc}$ inversion. Notice how this sends $T\to A'$, $M_A \to K$, and we'll call $Q'$ and $P'$ to the images of $Q$ and $P$ respectively. Clearly $P\neq Q'=M_BM_C \cap (ABC)$ and $P'=AQ'\cap BC$. So $(QPM_AK) \to (KM_AQ'P')$ and $PT \to (AA'P)$. We will show they both are tangent at $P'$, which will end the problem. Let $E$ be the antipode of $P'$ in $(KM_AQ'P')$. Clearly $\angle EM_AP'=90^\circ$, so $E$ lies on the perpendicular bisector of $BC$, which coincides with the perpendicular bisector of $AA'$. So we just need to show that $EA=EP'$ to end the problem. But this is triavial as $AQ'=Q'P'$ and $\angle AQ'E=\angle EQ'P'=90^\circ$ implies that triangles $AQ'E$ and $PQ'E$ are congruent, so we are done.

Let $A-$symedian intersect $(ABC)$ at $D$ and $T$ be the intersection of the tangent to $(ABC)$ at $A$ and $BC$. Claim: $P,Q,M_A,D$ are concyclic. Proof: If $S\in (ABC),AS\parallel BC,$ then \[(A,D;B,C)=-1=(SA,SM_a;SB,SC)=(A,SM_a\cap (ABC);B,C)\]thus, $S,M_a,D$ are collinear. \[\measuredangle DPQ=\measuredangle DSA=\measuredangle DM_AQ\]Which completes the proof.$\square$ Let $O$ be the circumcenter of $(ABC)$. Now invert around $(ABC)$. New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and midpoints of $BC,CA,AB$ are $M_a,M_b,M_c$ respectively. The ray $M_bM_c$ intersects $(ABC)$ at $P$ and $(OAP)$ meets $(OBC)$ at $Q$. $D$ is the intersection of $A-$symedian with $(ABC)$ and $T$ is the midpoint of $AD$. Prove that $(OPT)$ and $(PQD)$ are tangent to each other. We observe that $T$ is $A-$dumpty point. Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle A$. New Problem Statement: $ABC$ is a triangle with midpoints $M_a,M_b,M_c$. $K$ is the altitude from $A$ to $BC$ and the ray $M_cM_b$ intersects $(ABC)$ at $P$. If $D$ is the midpoint of $M_bM_c$ and $KP$ meets $(M_aM_bM_cK)$ at $Q,$ then show that $(PM_aK)$ and $(PQD)$ are tangent. Let $L$ be the reflection of $P$ with respect to $D$ and $H$ be the orthocenter of $\triangle ABC$. Let $X$ be the midpoint of $XP$. $H$ is the center of homothety sends the ninepoint circle to $(ABC)$ hence $N$ lies on $(M_aM_bM_cK)$.Let $N$ be the midpoint of $AH$. $AP\parallel NX$ and $PA=PK$ hence \[\measuredangle XNK=\measuredangle PAK=\measuredangle AKQ=\measuredangle NKQ\]And $X,Q,K,N$ are concyclic thus, $XQ\parallel NK$. This implies $Q$ is the midpoint of $KP$. Note that $PLKM_a$ is an isosceles trapezoid because the perpendicular to $M_bM_c$ at $D$ is the perpendicular bisector of both $KM_a$ and $LP$. If $l$ is the tangent line to $(PM_aK)$ at $P$, then \[\measuredangle PDQ=\measuredangle PLK=\measuredangle PM_aC=\measuredangle PKM_a+\measuredangle M_aPK=\measuredangle M_aPK+\measuredangle (l,PM_a)=\measuredangle (l,PQ)\]Thus, $l$ is tangent to $(PDQ)$ as desired.$\blacksquare$