Should say "$AM$ cuts $\Gamma_C$ at $E$" rather than "$AM$ cuts $C$ at $E$." Firstly, note that it's well-known that $D$ is the midpoint of the $A-$symmedian chord of $\triangle ABC.$ Therefore, we have that $\angle EAC = \angle MAC = \angle DAB = \angle ACD.$ Therefore, we have that $DE || AC.$ Therefore, as $ADEC$ is an isosceles trapezoid and $OA = OC$, we know that $OD = OE.$ Let $O_b, O_c$ be the centers of $\Gamma_C, \Gamma_B$, respectively (yes we switched them up). The following lemma is now helpful. Lemma. $O_bODO_c$ is an isosceles trapezoid. Proof. Observe that $OO_c \perp AB, AO_b \perp AB, AO_c \perp AC, OO_b \perp AC$ imply that $AO_bOO_c$ is a parallelogram. Therefore, we have that $OO_c = AO_b = DO_c$ and $OO_b = AO_c = DO_c.$ Together with $DO = OD$, we have that $\triangle O_cOD \cong \triangle O_bDO.$ This, together with $O_bO_c \perp AD, OD \perp AD$, gives the desired conclusion. $\blacksquare$ Let's now define $P = O_cD \cap O_bO.$ Since $P \in O_bO$, we have that $PD = PE.$ On the other hand, the lemma implies by symmetry that $PO = PD.$ Therefore, $P$ is actually the circumcenter of $\triangle ODE.$ Hence, since $P \in DO_c$, it's readily seen that $(\triangle ODE)$ is tangent to $\Gamma_B,$ as desired. $\square$

Pathological wrote: Should say "$AM$ cuts $\Gamma_C$ at $E$" rather than "$AM$ cuts $C$ at $E$." Firstly, note that it's well-known that $D$ is the midpoint of the $A-$symmedian chord of $\triangle ABC.$ Therefore, we have that $\angle EAC = \angle MAC = \angle DAB = \angle ACD.$ Therefore, we have that $DE || AC.$ Therefore, as $ADEC$ is an isosceles trapezoid and $OA = OC$, we know that $OD = OE.$ Let $O_b, O_c$ be the centers of $\Gamma_C, \Gamma_B$, respectively (yes we switched them up). The following lemma is now helpful. Lemma. $O_bODO_c$ is an isosceles trapezoid. Proof. Observe that $OO_c \perp AB, AO_b \perp AB, AO_c \perp AC, OO_b \perp AC$ imply that $AO_bOO_c$ is a parallelogram. Therefore, we have that $OO_c = AO_b = DO_c$ and $OO_b = AO_c = DO_c.$ Together with $DO = OD$, we have that $\triangle O_cOD \cong \triangle O_bDO.$ This, together with $O_bO_c \perp AD, OD \perp AD$, gives the desired conclusion. $\blacksquare$ Let's now define $P = O_cD \cap O_bO.$ Since $P \in O_bO$, we have that $PD = PE.$ On the other hand, the lemma implies by symmetry that $PO = PD.$ Therefore, $P$ is actually the circumcenter of $\triangle ODE.$ Hence, since $P \in DO_c$, it's readily seen that $(\triangle ODE)$ is tangent to $\Gamma_B,$ as desired. $\square$ sorry, can you please elaborate the well-known lemma? seems pretty foreign to me.. thankyou

Let $D'=DO \cap \Gamma_B $ and $K=AD\cap CE.$ Note that $AD' \perp AC.$ Clearly, $D$ lies on the $A$-symmedian of $\triangle{ABC} \implies$ $$\angle{CAD}=\angle{BAM}=\angle{ACE} \implies \angle{ACK}=\angle{CAK} \implies OK \perp AC$$Hence, because $AD'\mid \mid OK, $ by homothety it is enough to show that $ODKE$ is cyclic $$\angle{OME}=90-\angle{AMC}=90-\angle{B}-\angle{MAB}=\angle{OCA}-\angle{ACE}=\angle{OCE}\implies \angle{OEK}=90$$Because $\angle{ADC}=\angle{ADB}=180-\angle{A},$ we have that $AD$ is the internal bisector of $\angle{CDB}$ and $CDOB$ is cyclic, and since $O$ is on the bisector of $BC$ this means that $DO$ is the external bisector of $\angle{CDB},$ implying that $\angle{KDO}=\angle{ADO}=90 \ \ \ \blacksquare$

well-known that $D$ is the $A$-dumpty point so $ODBC$ is cyclic and$AD$ is $A$-symmedian in $\triangle ABC$ let $F=BD \cap AM $ and $CE \cap (ABC) =C'$ cliam(1):$OEMC$ is cyclic proof: $\angle C'BA=\angle C'CA=\angle EAB$ so $C'B \parallel AM $ so $\angle OEC=90$ $\blacksquare$ claim(2): $DFOE$ is cyclic proof: $\angle FDO=\angle OCB=\angle OEF$ $\blacksquare$ $\angle BAD=\angle BCO=\angle FEO $ this will imply that the two circles are tangent

Let $K , L$ be the midpoint of $AC , AB , F$ the projection of $A$ on $BC , J$ the intersection of $ML$ with the $A$-symmedian Consider an inversion around $A$ with power of $\sqrt{AB.AC}$ followed by a reflection over the $\hat{A}$ bisector then $B$ and $C$ swap and $AM , \Gamma_B$ and $\Gamma_C$ map to the $A$-symmedian , a line through $C$ parallel to $AB$ , and a line through $B$ parallel to $AC$ respectively so $D$ maps to the reflection of $A$ over $M , O$ maps to the reflection of $A$ over $BC$ (well-known lemma) , $E$ maps to the intersection of the $A$-symmedian with the line through $B$ parallel to $AC$ take a dilation at $A$ with ratio $0.5$ and consider the images of these under the last three transformations: $O \mapsto F$ $D \mapsto M$ $E \mapsto J$ $\Gamma_B \mapsto MK$ $\angle{LAM}=\angle{CAJ}=\angle{AJM}$ So $LF^2=LA^2=LM.LJ$ Hence $\angle{KMF}=\hat{B}=\angle{LFM}=\angle{MJF}$ implying that $MK$ tangents $(MJF)$ as desired

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