I think it is equialent with to prove that $\angle ADB=\angle BDM$. It is obvious that $\angle ABC=\angle BMD$ and $\angle ACB=\angle DMC$. If We say $\angle BAD=\alpha$ and $\angle CBM=\beta$ By Lemma in triangle $ABC$ $\to$ \[\frac{\sin ABC}{\sin ACB}=\frac{\sin \alpha}{\sin \alpha+\sin \angle ABC+\angle ACB}\]And in triangle by the same lemma \[\frac{\sin BMD}{\sin DMC}=\frac{\sin ABC}{\sin ACB}=\frac{\sin \beta}{\sin \beta+\angle BMD+\angle DMC}\]Combining this two equations we get $\alpha = \beta$ Then in In triangle $ABD$ we get $\angle ADB=180^{\circ}-\alpha-\angle ABC$ and in the triangle $BMD$ , $\angle BDM=180^{\circ}-\alpha-\angle BMD=180^{\circ}-\alpha-\angle ABC$.$\to \angle ADB=\angle BDM$

Note that BMC=BMD+CMD=B+C=180-A, so M lies on (ABC) Consider the transformation P: inversion at A with power AB*AC, and reflecting it across the angle bisector of BAC. Let P(X) be the image of X under this transformation. We claim that P(D)=M. Let X, Y be the intersection of the angle bisector of BAC with BC and (BAC). Then DMY=DMC+CMY=C+XAC=DXY, so M lies on (DXY), with diameter XY. But P(X)=Y and P(Y)=X, so P(circle DXY)=circle DXY because the diameter XY is unchanged in the transformation. So P(D)=M, so since AD is the median, AM is the symedian. Now it is not hard. If AD intersect (ABC) at E, and F is reflection of E about D, then we claim that M'=F. Note that BFC=BEC=BMC=BM'C, and M'BC=MBC=ECB=FBC (because ME parallel BC and FBEC is parallelogram), so M'=F, and we are done.

We can easily prove that $M$ lies on the A-symmedian. We have: $\angle BMC = \angle BMD + \angle DMC = \angle ABC + \angle ACB = 180 - \angle BAC$. Hence $M$ lies on the $\odot (ABC)$. But now we have: $\angle DMC = \angle BCA = \angle BMA$. From this, since $D$ is the mid-point of $BC$ we have that $AM$ is the M-symmedian of $\triangle BMC$. But since $ABMC$ is cyclic it's well-known (and easy to prove) fact that $AM$ is also a A-symmedian of $\triangle ABC$. Now a simple angle chase gives us: $\angle BDA = \angle ACB + \angle DAC = \angle AMB + \angle MAC = \angle FBM = \angle BDM = \angle BDM'$, where $F$ is a point on $AB$, s.t. $B$ is between $A$ and $F$. Now from this it's obvious that $D-M'-A$ are colinear. Q.E.D.

My solution: Lemma: Let $\triangle ABC$ be a triangle and let $M$ is the midpoint of $BC$ $\Longrightarrow$ $\sin\angle ABC.\sin\angle MAC=\sin\angle ACB.\sin\angle MAB$ Proof (It's trivial by law of sines in $\triangle BAM$ and $\triangle MAC$) In the problem: It is easy to see that:$\angle BMD=\angle ABC$ and $\angle DMC=\angle BCM$ $\Longrightarrow$ By lemma we get: $\sin\angle MBC.\sin\angle DMC=\sin\angle MCB.\sin\angle DMB...(1)$ and $\sin\angle ABC.\sin\angle DAC=\sin\angle ACB.\sin\angle DAB...(2)$ $\Longrightarrow$ By $...(1)$ and $...(2)$ we get $\angle BAD=\angle CBM$ and $\angle CAD=\angle BCM$ and let $M_1$ the reflection of $M'$ in $D$ $\Longrightarrow$ $ABCM_1$ are concyclic $\Longrightarrow$ $M',A,M_1,D$ are collinear $\Longrightarrow$ $M'$ is on $AD$...

A proof without any additional point Note that $\widehat{BMD}=\widehat{ABD},\widehat{DMC}=\widehat{ACD}$ $\Rightarrow\widehat{BMC}=\widehat{ABC}+\widehat{ACB}=180-\widehat{BAC}$ $\Rightarrow M\in (ABC)$. $\Rightarrow\widehat{AMC}=\widehat{ABC}=\widehat{ABD}=\widehat{BMD}$ $\Rightarrow$ triangle $AMC$ are similar to triangle $BMD$ and $\widehat{AMB}=\widehat{DMC}$ $\Rightarrow\frac{MC}{MD}=\frac{AC}{BD}=\frac{AC}{CD}$ Hence triangle $ACD$ are similar to triangle $CMD$, which means $\widehat{ADC}=\widehat{MDC}$. $\Rightarrow AD,AM$ are symmetric wrt $BC$. $\Rightarrow M'\in AD$.

This problem is so cool!!!

We claim that $M'$ is the $A$ Humpty point. Indeed let $K$ be the point where the $A$ symmedian cuts $(ABC)$ and $N$ be the midpoint of $BC$ Then Since $KA$ is the $K$ symmedian in triangle $KBC \implies \angle NKB = \angle CKA = \angle CAB = \angle NBA \implies (NKB) $ is tangent to $AB$.Similarly We obtain the other tangency also.Hence we may conclude that $K=M$.And therefore $M'$ lies on the $A$ median.

$\angle DMC=\angle BCA, \! \angle BMD=\angle ABC \Rightarrow$$\angle BMC=\angle BMD+\angle DMC=$ $=\angle ABC+\angle ACB=180^{\circ}-\angle BAC\Rightarrow$$A,B,M,C$ are concyclic. Let $T$ be the reflection of $M$ over $D$. Then $BTCM$ is a parallelogram and $\angle BTM = \angle TMC = \angle ACB$, $\angle BMT = \angle ABC \Rightarrow \triangle BTM \sim \triangle ABC \Rightarrow \angle DBT = \angle DCM = \angle DAC$. Let $U$ be the reflection of $M^{\prime}$ over $D$. $BM^{\prime}CU$ is a parallelogram $\Rightarrow \angle M^{\prime}CD= \angle MCD = \angle UBD$. Let's notice that $DH \parallel UM$ where $H=MM^{\prime}\cap BC$ as $D,H$ are midpoints of $UM^{\prime}, MM^{\prime}$ respectively. Thus, $BC\parallel UM$ and $BUMC$ is an isosceles trapezoid. Thus, $B,U,M,C$ are concyclic. It implies that $A,B,U,M,C$ are concyclic with $\angle UBC = \angle UAC$. But we proved that $\angle UBC=\angle UBD =\angle MCD = \angle DAC$. Thus, $\angle UAC = \angle DAC$ that means collinearity of $A,D,U$. But $M^{\prime}$ lies on $UD$, so $M^{\prime}$ lies on $AD$ as desired.

Note that $D$ lies on $(ABC)$ as $$ \measuredangle CBA = \measuredangle DMB $$and $$ \measuredangle ACB = \measuredangle CMD$$, so add this both to get $D \in (ABC)$. Now take a $\sqrt{bc}$ inversion centered at $A$ , followed by reflection around the angle bisector of $ \angle A$ ,note that the pair ${(M, D)}$ will be swapped under this mapping , so as $AD$ is median , it follows that $AM$ is $A$ symmedian. Now as the reflection of $M$ around $BC$ will be $A$ $\text{Humpty Point}$ and it will obviously lies on $A$ median.

twotimestwo wrote: Let $ABC$ be a triangle and $D$ be the midpoint of the segment $BC$. The circle that passes through $D$ and tangent to $AB$ at $B$, and the circle that passes through $D$ and tangent to $AC$ at $C$ intersect at $M\neq D$. Let $M'$ be the reflection of $M$ with respect to $BC$. Prove that $M'$ is on $AD$. Let $X$ be the point on the $A$-Symmedian distinct from $A$ lying on the circumcircle of $\triangle ABC$. We see that $\angle BXD = \angle AXD + \angle AXB = (180 - \angle BAC - 2\angle BCA) + \angle BCA = 180 - \angle BAC - \angle BCA = \angle ABC = \angle ABD \implies \odot(\triangle BXD)$ is tangent to line $\overset{\longrightarrow}{AB}$ and similarly $\odot(\triangle CXD)$ is tangent to line $\overset{\longrightarrow}{AC}$, which means that $X \equiv M$. It is well-known that the reflection of $X$ about line $\overset{\longrightarrow}{BC}$ is the $A$-Humpty Point which lies on the $A$-Median, as desired.