Let $ H $ be the orthocenter of $ \triangle ABC, $ $ \triangle H_aH_bH_c $ be the orthic triangle of $ \triangle ABC $ and let $ Y,Z $ be the intersection of $ OD $ with $ CA, AB, $ respectively. Clearly, $ \triangle AHH_b \stackrel{-}{\sim} \triangle YCD $ and $ \frac{AE}{HE} = \frac{DF}{OF} = \frac{DL}{CL} = \frac{YN}{CN}, $ so $ \triangle EHH_b \stackrel{-}{\sim} \triangle NCD $ and $$ \angle ECB = \angle EH_bH = \angle CDN. $$Similarly, $ \angle EBC = \angle EH_cH = \angle BDM, $ so $ DM \perp DN. $ Furthermore, from $ DK = DL $ we get $ OD $ is the $ D $-median of $ \triangle DMN, $ so $ \odot (DMN) $ is tangent to $ BC $ at $ D. $ Let $ T $ be the Miquel point of $ D,M,N $ WRT $ \triangle ABC, $ then $$ \angle BTC = 270^{\circ} - \angle MTN = 90^{\circ} + \angle BAC, $$so $ \odot (BTC) $ is tangent to $ BO, CO $ at $ B,C, $ respectively $ \Longrightarrow $ $ \odot (BTC) \equiv \omega. $ Finally, note that $$ \angle CBT + \angle TMN = \angle DMN = \angle CDN = \angle CTN, $$so $ \odot (AMN) $ is tangent to $ \omega $ at $ T. $

But this problem is mine , please the link http://forumgeom.fau.edu/FG2014volume14/FG201425.pdf

Let $H$ be the orthocenter of $\triangle ABC$ and let $U$ be the antipode of $A$ in $(O) \equiv \odot(ABC).$ If $BU,CU$ cut $AC,AB$ at $Y,Z,$ then $U$ is orthocenter of $\triangle AYZ$ $\Longrightarrow$ $\omega$ is the circle with diameter $YZ.$ If $OD$ cuts $AB,AC,$ at $P,Q,$ then clearly $\overline{PBZ} \sim \overline{QCY}$ $\Longrightarrow$ $(O),$ $\odot(AYZ)$ and $\odot(APQ)$ meet at $A$ and the center $R$ of the spiral similarity that swaps $\overline{BZ}$ and $\overline{CY}.$ Furthermore, $(O)$ and $\odot(APQ)$ are orthogonal, due to $\angle APQ=90^{\circ}-\angle ABC=\angle OAQ.$ $AR,BC,YZ$ are pairwise radical axes of $\omega,(O),\odot(AYZ)$ concurring at their radical center $T$ and $AO$ is the polar of $T$ WRT $\omega.$ Thus if $AO$ cuts $\omega$ at $S$ (inside of $\triangle ABC$), then $TS$ is tangent of $\omega$ $\Longrightarrow$ $TS^2=TB \cdot TC=TA \cdot TR$ $\Longrightarrow$ $\odot(ARS)$ is tangent to $\omega$ at $S.$ If $\odot(ARS)$ cuts $\overline{AB},\overline{AC}$ at $M',N',$ then the ratio of the powers of $M',N',S$ WRT $\odot(APQ)$ and $(O)$ are equal $\Longrightarrow$ $\frac{M'A \cdot M'P}{M'A \cdot M'B}=\frac{N'A \cdot N'Q}{N'A \cdot N'C}=\frac{SA^2}{SA \cdot SU} \Longrightarrow \frac{PM'}{M'B}=\frac{QN'}{N'C}=\frac{AS}{SU} \ \ (1).$ But on the other hand, keeping in mind that $\triangle ABC \cup E \sim \triangle AYZ \cup S,$ we have $\frac{AS}{SU}=\frac{AE}{EH}=\frac{DF}{FO}=\frac{DL}{LC}=\frac{DK}{KB}=\frac{PM}{MB}=\frac{QN}{NC} \ (2).$ From $(1)$ and $(2),$ we get $M \equiv M',N \equiv N'$ $\Longrightarrow$ $\odot(AMN)$ touches $\omega$ at $S.$

Here is a length bash solution I found - originally I tried to find various lengths as to use Casey's Theorem, however when I noticed that $ MK + NL = MN $ the solution fell into place. For the remainder of the proof, let $ a = BC, b = CA, c = AB, $ and $ R $ be the circumradius of $ \triangle{ABC}. $ Lemma 1: $ MK + NL = MN. $ Proof: Since $ A, G, D $ are collinear and $ OD \parallel AE $ we find that $ \frac{AE}{FD} = \frac{AG}{GD} = 2. $ Now let $ X $ projection from $ A $ onto $ BC. $ Looking at the Power of $ X $ with respect to the circle with diameter $ BC, $ we find that $ EX = \sqrt{BX * CX}. $ Now it is easy to compute that $ AX = b\sin{C} $ and $ BX = c\cos{B} $ and $ CX = b\cos{C} $ so we find that: \[ FD = \frac{AE}{2} = \frac{AX - EX}{2} = \frac{b\sin{C} - \sqrt{bc\cos{B}\cos{C}}}{2} \] Now it is easy to compute that $ OD = R\cos{A}. $ Therefore we have that $ \frac{DK}{FD} = \frac{DB}{OD} $ so we find that: \[ CL = BK = DB - DK = DB - \frac{DB \cdot FD}{OD} = \frac{a}{2} - \frac{a \cdot FD}{2OD} = \frac{a}{2} - \frac{ab\sin{C} - a\sqrt{bc\cos{B}\cos{C}}}{4R\cos{A}}\hspace{5 mm}(1) \] Now note that $ MK = BK\tan{B} $ and $ NL = BK\tan{C} $ so $ MK + NL = BK(\tan{B} + \tan{C}) = \frac{BK\sin{A}}{\cos{B}\cos{C}}. $ Moreover note that $ AM = c - MB = c - \frac{BK}{\cos{B}} $ and similarly $ AN = b - \frac{BK}{\cos{C}}. $ Therefore by the Law of Cosines on $ \triangle{AMN} $ we find that \[ MN^2 = \left(c - \frac{BK}{\cos{B}}\right)^2 + \left(b - \frac{BK}{\cos{C}}\right)^2 - 2\left(c - \frac{BK}{\cos{B}}\right)\left(b - \frac{BK}{\cos{C}}\right)\cos{A} = \] \[ a^2 + \left(\frac{2b\cos{A}\cos{C} + 2c\cos{A}\cos{C} - 2b\cos{B} - 2c\cos{C} }{\cos{B}\cos{C}}\right)BK + \left(\frac{\cos^2{B} + \cos^2{C} - 2\cos{A}\cos{B}\cos{C}}{\cos^2{B}\cos^2{C}}\right)BK^2 \] So we want $ BK $ to be the positive solution to: \[ a^2 + \left(\frac{2b\cos{A}\cos{C} + 2c\cos{A}\cos{C} - 2b\cos{B} - 2c\cos{C} }{\cos{B}\cos{C}}\right)x + \left(\frac{\cos^2{B} + \cos^2{C} - 2\cos{A}\cos{B}\cos{C}}{\cos^2{B}\cos^2{C}}\right)x^2 = \left(\frac{1 - \cos^2{A}}{\cos^2{B}\cos^2{C}}\right)x^2 \] Using the facts that $ a\cos{A} + b\cos{B} - c\cos{C} = 2c\cos{A}\cos{B} $ (and cyclic) and $ \cos^2{A} + \cos^2{B} + \cos^2{C} + 2\cos{A}\cos{B}\cos{C} = 1 $ we have that this quadratic simplifies to: \[ \left(4\cos{A}\right)x^2 + \left(4a\cos{B}\cos{C}\right)x - a^2\cos{B}\cos{C} = 0 \] Which by the quadratic formula and the well-known identities $ AX = 2R(\cos{A} + \cos{B}\cos{C}) $ has solution: \[ \frac{-a\cos{B}\cos{C} + a\sqrt{\cos{B}\cos{C}(\cos{A} + \cos{B}\cos{C})}}{2\cos{A}} = \] \[ \frac{a(A2R\cos{A}- AX) + a\sqrt{2R\cos{B}\cos{C} \cdot AX}}{4R\cos{A}} = BK \] So the lemma is proven. Now, returning to the original problem, let $ Y $ be the point on $ MN $ such that $ MY = MK $ and $ NY = NL. $ Let the rotation centered at $ M $ that takes $ K $ to $ Y $ also take $ B $ to the point $ Z. $ Note that the rotation centered at $ N $ that takes $ L $ to $ Y $ also takes $ C $ to $ Z. $ Lemma 2: Quadrilateral $ AMZN $ is cyclic. Proof: Note that $ \angle{MZN} = \angle{MZY} + \angle{NZY} = \angle{B} + \angle{C} = 180 - \angle{A} $ so the lemma is proven. Lemma 3: $ Z \in \omega. $ Proof: $ \angle{BZC} = 360 - \angle{BZM} - \angle{MZN} - \angle{CZN} = 360 - \left(\frac{180 - \angle{BMZ}}{2}\right) - (180 - \angle{A}) - \left(\frac{180 - \angle{CNZ}}{2}\right) = \angle{A} + \frac{\angle{ANZ} + \angle{AMZ}}{2} = \angle{A} + 90 $ so the lemma is proven. Lemmas 2 and 3 indicate that $ Z $ is one of the intersections between $ \omega $ and the circumcircle of $ \triangle{AMN}. $ Letting $ R $ be the center of the circumcircle of $ \triangle{AMN} $ and $ S $ be the center of $ \omega $ it suffices to show that $ R, Z, S $ are collinear. But $ \angle{RZM} + \angle{MZB} + \angle{BZS} = (90 - \angle{ZNM}) + \angle{MBZ} + (90 - \angle{ZCB}) = \angle{C} + (\angle{B} - \angle{ZBC}) + (90 - \angle{ZCB}) = \angle{B} + \angle{C} + \angle{A} = 180 $ so we are done. Phew!

Just wanna point out that verifying $MK+NL = KL$ can be less computational. It seems to me that once you've found the relation, it's fairly easy to proceed to what we seek.AThe main reason is that one may find it intuitive to rotate $MK$ around $M$ and $NL$ around $N$ until they meet on $MN$, and see what the intersection point gets you from there. However, speculating it first-handed is not so obvious, at least not to me anyway. Now we shall prove it! One can see that $(MK-NL)^2+KL^2= MN^2$, due to Pythagoras' Theorem.So it suffices to show that $$KL^2=4MK*NL.$$ Clearly, $BK=LC$ and $KD=DL$. So for our convenience, let'es denote them by $x$ and $y$ respectively. Now we have $MK=x*tanB$ and $NL=x*tanC$, and obviously $KL=2y$. Hence, our goal has now become showing that $$y^2=x^2*tan(B)tan(C).$$ Let $X$ be the foot of the alitude from $A$ on $BC$. Then it's easy to see that $BX*XC=EX^2$.Now one can easily obtain that $BX=AX/tan(B),XC=AX/tan(C)$.Therefore, $\dfrac{AX^2}{tanB*tanC}=EX^2$, which implies $tanB*tanC= (AX/EX)^2$. Again, The result we seek has now mutated to proving $$AX:EX= y:x. $$ This seems simple enough to prove synthetically.I ,however, still have to rely on some more trigonometrical notations, although it's still short and concise. Since $FD=AE/2$, we have that $y=DL=\frac{1}{2}*AE*tan(A)$ ,and then $x=\frac{1}{2}*(a-AE*tanA)$.Moreover, we have $EX=AX-AE$.Combining all these, we obtain the ultimate relation of what we have to prove $$\frac{AX}{AX-AE}=\frac{AE*tan(A)}{a-AE*tan(A)}.$$ Now Multiplying the denominators up, rearranging them a bit, we have that $$AX*a=AE*tanA*(2AX-AE).$$ Observe that $AX*a= AC*ABsinA=2*[ABC]$, thus $AX*a=AC*ABcosA*tanA=AC*AH_b*tan(A)$,where $H_b$ is obviously the foot of altitude from $B$ on $CA$.Hence, we finally obtain $$AC*AH_b= AE*(2h_a-AE).$$ However, this is merely just the power of $A$ wrt. the circle passing through $B,C$,centred at $D$.

Let $H$ orthocenter of $\triangle ABC$, and $ Y \equiv BH \cap AC, X \equiv OD \cap AC$ Now $\frac{AE}{EH}=\frac{DF}{FO}=\frac{DL}{LC}=\frac{XN}{NC}$, then $\triangle XDC \cup N \sim \triangle AYH \cup E \Rightarrow \angle HYE=\angle NDC=\angle ECB$ analogously $\angle MDB=\angle EBC \Rightarrow MDN=90^{\circ}$, if $P \equiv MN \cap OD \Rightarrow MP=NP=PD$ (Because $KD=DL$) Then $\angle DNM=\angle MDB$ Let $Z$ the miquel point of $D,N,M$ WRT $\triangle ABC$ (i.e. $Z \equiv \odot (BMD) \cap \odot (MAN) \cap \odot (NDC)$ ) $\angle BZC =\angle BMD +\angle CND = 90^{\circ}+ \angle BAC$, easy see that $Z \in \omega$, $\angle BZM=\angle MND=\angle MNZ+\angle DNZ =\angle MNZ + \angle ZCB$ Notice $Z \in \omega, Z \in \odot (AMN)$ and $\angle BZM=\angle MNZ + \angle ZCB \Rightarrow \omega$ is tangent to $\odot (AMN)$

Luis González wrote: Let $H$ be the orthocenter of $\triangle ABC$ and let $U$ be the antipode of $A$ in $(O) \equiv \odot(ABC).$ If $BU,CU$ cut $AC,AB$ at $Y,Z,$ then $U$ is orthocenter of $\triangle AYZ$ $\Longrightarrow$ $\omega$ is the circle with diameter $YZ.$ If $OD$ cuts $AB,AC,$ at $P,Q,$ then clearly $\overline{PBZ} \sim \overline{QCY}$ $\Longrightarrow$ $(O),$ $\odot(AYZ)$ and $\odot(APQ)$ meet at $A$ and the center $R$ of the spiral similarity that swaps $\overline{BZ}$ and $\overline{CY}.$ Furthermore, $(O)$ and $\odot(APQ)$ are orthogonal, due to $\angle APQ=90^{\circ}-\angle ABC=\angle OAQ.$ $AR,BC,YZ$ are pairwise radical axes of $\omega,(O),\odot(AYZ)$ concurring at their radical center $T$ and $AO$ is the polar of $T$ WRT $\omega.$ Thus if $AO$ cuts $\omega$ at $S$ (inside of $\triangle ABC$), then $TS$ is tangent of $\omega$ $\Longrightarrow$ $TS^2=TB \cdot TC=TA \cdot TR$ $\Longrightarrow$ $\odot(ARS)$ is tangent to $\omega$ at $S.$ If $\odot(ARS)$ cuts $\overline{AB},\overline{AC}$ at $M',N',$ then the ratio of the powers of $M',N',S$ WRT $\odot(APQ)$ and $(O)$ are equal $\Longrightarrow$ $\frac{M'A \cdot M'P}{M'A \cdot M'B}=\frac{N'A \cdot N'Q}{N'A \cdot N'C}=\frac{SA^2}{SA \cdot SU} \Longrightarrow \frac{PM'}{M'B}=\frac{QN'}{N'C}=\frac{AS}{SU} \ \ (1).$ But on the other hand, keeping in mind that $\triangle ABC \cup E \sim \triangle AYZ \cup S,$ we have $\frac{AS}{SU}=\frac{AE}{EH}=\frac{DF}{FO}=\frac{DL}{LC}=\frac{DK}{KB}=\frac{PM}{MB}=\frac{QN}{NC} \ (2).$ From $(1)$ and $(2),$ we get $M \equiv M',N \equiv N'$ $\Longrightarrow$ $\odot(AMN)$ touches $\omega$ at $S.$ Why the ratio of power of M',N',S (APQ),(O) equal

Double-posting because a decent proof exists Though it is probably equivalent to Telv's proof. TelvCohl wrote: Let $ \triangle ABC $ be an acute triangle with circumcircle $\gamma$ with center $ O $ and centroid $ G$. Let $ D $ be the midpoint of $ BC $ and $ E\in \odot (BC) $ be a point inside $ \triangle ABC $ such that $ AE \perp BC$. Let $ F=EG \cap OD $ and $ K, L $ be the point lie on $ BC $ such that $ FK \parallel OB, FL \parallel OC$. Let $ M \in AB $ be a point such that $ MK \perp BC $ and $ N \in AC $ be a point such that $ NL \perp BC$. Let $ \omega $ be a circle tangent to $ OB, OC $ at $ B, C, $ respectively. Prove that the circumcircle $\Gamma$ of $\triangle AMN$ is tangent to $ \omega $. Let $\overline{AA'}$ be the diameter of $\gamma$ and $S \in \gamma$ with $BCA'S$ harmonic. Let $T=\overline{AS} \cap \overline{BC}$, $L=\overline{AA'} \cap \overline{BC}$ and $X=\overline{AA'} \cap \omega$ with $X$ closer to point $A$. Note $\left(\overline{BC}, \overline{LT}\right)=-1$. Lemma: $\Gamma$ passes through $S$. (Proof) Notice that $\measuredangle SBM=\measuredangle SCN$ and $$\frac{BM}{CN}=\frac{BK \cdot \tfrac{1}{\cos B}}{CL \cdot \tfrac{1}{\cos C}}=\frac{\cos C}{\cos B}=\frac{BA'}{CA'}=\frac{BS}{CS}$$proving $\triangle SBM \sim \triangle SCN$ hence $S$ lies on $\Gamma$. Remark that the proof ignores $AE=2DF$ and thus holds for all points $F$ on line $\overline{DO}$. $\blacksquare$ Lemma: Line $\overline{TX}$ is tangent to $\omega$. (Proof) Note that $T$ lies on the polar $\overline{BC}$ of $O$ in $\omega$; while $\left(\overline{BC}, \overline{LT}\right)=-1$ forces $L$ to lie on the polar of $T$ in $\omega$. Hence $\overline{OL}$ coincides with the polar of $T$ in $\omega$; proving the claim. $\blacksquare$ Lemma: $\tfrac{DF}{DO}=\tfrac{AX}{AA'}$ (Proof) Let $\overline{AE}$ meet $\odot(BC)$ again at $U$; let $H$ be the orthocenter of $\triangle ABC$ and $J$ be the foot of $A$-altitude. Notice that $A$ and $H$ are inverses in $\odot(BC)$ hence $\left(\overline{EU}, \overline{HA}\right)=-1$. Thus, $AE \cdot AU=AH \cdot AJ=bc\cos A$. Observe that $\angle BXC=90^{\circ}+\angle A$ hence a $\sqrt{bc}$ inversion maps point $U$ to point $X$. Thus, $AU \cdot AX=AB \cdot AC$ so $AE=AX \cos A$. Now $$\frac{DF}{DO}=\frac{AE}{AH}=\frac{AX \cdot \cos A}{AA' \cos A}$$proving the claim. $\blacksquare$ Lemma: Point $X$ lies on circle $\Gamma$. (Proof) Let $\Gamma$ meet $\overline{AA'}$ at points $A$ and $R$. Move point $F$ along $\overline{DO}$ uniformly. Spiral similarity at $S$ shows that $R$ also moves uniformly. Now if $F=D$ we see $R=A$ while if $F=O$ we see $R=A'$. Since $\tfrac{DF}{DO}=\tfrac{AX}{AA'}$ we conclude that $R=X$ when $\overrightarrow{AE}=-2\overrightarrow{DF}$ as desired. $\blacksquare$ Suppose $\Gamma, \omega$ meet at $X,Y$. Apply radical axis theorem to $\Gamma, \gamma, \omega$ we conclude that lines $\overline{AS}, \overline{BC}, \overline{XY}$ concur at point $T$. Hence $X=Y$ proving the desired tangency. Our proof is complete! $\blacksquare$.

Let the Miquel point of D,M,N be X, H is the orthocentre, OD cuts AC at F Note that FK//OB, FL//OC => BK=LC => MNLK is a trapezoid so midpoint I lies on OD By thales theorem AE/EH=FD/FO=LD/LC=NC/NP We observe that AHQ ~ PCD so EHQ~NCD which means EQH=EQB=ECB=NDC Similarly, EBC=MDB so MDN=BEC=90 so IN=IM=ID (MDN is right-angle) Therefore BXC=360-MXB-MXN-NXC=180+A-MDB-NDC=90+A hence X lies on (T) Last step, we have MXB=MDB=90-NDC=DNL=IDN=IND=MNX+XND=MNX+XCD By tangent criterion, this means (AMN) is tangent to (T)

Any inversion solution for this ??

TelvCohl wrote: Let $ H $ be the orthocenter of $ \triangle ABC, $ $ \triangle H_aH_bH_c $ be the orthic triangle of $ \triangle ABC $ and let $ Y,Z $ be the intersection of $ OD $ with $ CA, AB, $ respectively. Clearly, $ \triangle AHH_b \stackrel{-}{\sim} \triangle YCD $ and $ \frac{AE}{HE} = \frac{DF}{OF} = \frac{DL}{CL} = \frac{YN}{CN}, $ so $ \triangle EHH_b \stackrel{-}{\sim} \triangle NCD $ and $$ \angle ECB = \angle EH_bH = \angle CDN. $$Similarly, $ \angle EBC = \angle EH_cH = \angle BDM, $ so $ DM \perp DN. $ Furthermore, from $ DK = DL $ we get $ OD $ is the $ D $-median of $ \triangle DMN, $ so $ \odot (DMN) $ is tangent to $ BC $ at $ D. $ Let $ T $ be the Miquel point of $ D,M,N $ WRT $ \triangle ABC, $ then $$ \angle BTC = 270^{\circ} - \angle MTN = 90^{\circ} + \angle BAC, $$so $ \odot (BTC) $ is tangent to $ BO, CO $ at $ B,C, $ respectively $ \Longrightarrow $ $ \odot (BTC) \equiv \omega. $ Finally, note that $$ \angle CBT + \angle TMN = \angle DMN = \angle CDN = \angle CTN, $$so $ \odot (AMN) $ is tangent to $ \omega $ at $ T. $ You can prove your claims much easier without using $Y , Z , Ha , Hb ,Hc$ . for proof that $T$ (Miquel point of $D , M , N$) is on $\omega$ , it's enough to say $\angle{MDN}=90$ because we can easily find that $\angle{BPC}=\angle{A}+90$ and it's the thing that we want to proof $T$ is on $\omega$ . For proof that it's enough to say $\triangle{DNL}$ and $\triangle{MDK}$ are similar (1) so we should say $DK.DL=LN.MK$ and it's easy to proof because $DF.AE=EH.FO$ and $DL=DK$ and $DK.OF=BK.DF$ . For proof that the two circles are tangent we just need to say $\angle{PBC}+\angle{PMN}=\angle{NPC}$ it means that we should just proof that $\angle{NMD}=\angle{NDC}$ and for that it's enough to show that $\triangle{DLM}$ is similar to $\triangle{MDN}$ , and it's easy because from (1) we know $NL.MD=DN.DK=DN.DL$ . And the proof finished .

Solved with naman12 and awang11 Let $E_1E_2E_3$ be the orthic triangle of $\triangle ABC$. Claim: $\angle NDM = 90^\circ$. Proof: $\angle NDM = 90^\circ \iff KD^2 = MK\cdot NL \iff MK = BK\tan B \iff (\frac{KD}{BK})^2 = \tan B \tan C$. SInce $KD/BK = FD/OF = AE/HE$ where $O,H$ are the circumcenter and orthocenter of $\triangle ABC$ respectively, it is equivalent to verify that $AE/HE = \sqrt{\tan B \tan C}$. From $AE(AE + 2EE_1) = AE_3\cdot AB = bc\cos A$ and $EE_1^2 = BE_1 \cdot E_1C = bc\cos B \cos C$ we deduce that $AE = R(2\sin B \sin C - \sqrt{\sin 2B \sin 2C})$ where $R$ is the circumradius. We also have $HE = 2R\cos A - 2R\sin B \sin C + R\sqrt{\sin 2B \sin 2C} \implies \frac{AE}{HE} = \frac{2\sin B \sin C - 2\sqrt{\sin B \sin C \cos B \cos C}}{2\sqrt{\sin B \sin C - \cos B \cos C}-2\cos B \cos C} = \sqrt{\tan B \tan C}$ as desired. $\blacksquare$ Define $I$ to be the reflection of $L$ over $DN$, then $I$ is also the reflection of $K$ oer $DM$ from the above lemma. Now let $P \in \overline{DI}$ s.t. $MNDP \sim CBEH$, then $AMNP$ cyclic and $\angle BAO = \angle HE_3E_2 =\angle MNP = \angle MAP \implies A,O,P$ collinear. SInce $\angle NPI = \angle ACB$ we have $DPNC$ cyclic, similarly $DPMB$ cyclic. Ntoe that $(BPC)$ tangent to $OB$ iff $2(\pi -\angle BPC) = \pi - \angle BOC \iff \pi = 2\angle BPC - \angle BOC$ which would follow from $EE_2E_2 \sim PBC$, a claim that would follow from the fact that $E,P$ are isogonal conjugates from $\triangle ABC$. To see that $E,P$ are isogonal conjugates, note that $\angle NCP = \angle NDP = \angle BEE_1 = \angle ECB$, as desired. Now we show that $P$ is the desired tangency point in the problem. This is equivalent to $|angle NMP = \angle CPN - \angle PBC$ which is true iff $\angle NAO = \angle CDN - \angle PBC$, i.e. $\angle NMP = \angle PBC - \angle NPC$ $\iff 90^\circ - B = \angle NDP - \angle PMD \iff 90^\circ - B = \angle BEH - \angle HCE$ but $\angle BEH - \angle HCE = \frac{\widehat{BE} - \widehat{EE_3}}{2} = \frac{\widehat{EE_3}}{2} = 90^\circ - B$, so we are done.

Let $P$ be the projection of $A$ on $BC$. Let $Y$ be the intersection of $(BMD)$ and $(NCD)$. By Miquel's theorem it lies on $(AMN)$. We now show that it lies on $\omega$ Indeed we have $\angle BYC=\angle BMD+\angle DNC=180^{\circ}-A+180^{\circ}-\angle MDB+\angle NDC$. Our desired result is $\angle BYC=90^{\circ}+A$. Hence we only need $\angle MDN=90^{\circ}$. Now notice that if $H$ is the orthocenter of $\triangle ABC$ then $$\frac{\tan\angle MDB}{\tan B}=\frac{BK}{KD}=\frac{OF}{FD}=\frac{HE}{AE}$$similarly $$\frac{\tan\angle NDC}{\tan C}=\frac{HE}{AE}$$Since we want to show that $\tan\angle MDB\tan\angle NDC=1$ it suffices to prove $$(\frac{AE}{HE})^2=\tan B\tan C$$We see that the R.H.S. is equal to $$\frac{PA^2}{PB\times PC}=\frac{PA^2}{PE^2}$$Moreover we have $PE^2=PB\times PC=PA\times PH$ Hence $$1-\frac{PH}{PE}=1-\frac{PE}{PA}$$$$\frac{HE}{PE}=\frac{AE}{AP}$$Therefore $$\frac{AE}{HE}=\frac{PA}{PE}$$as desired. This shows that $Y$ lies on $\omega$. Now since $M$ and $N$ are symmetric about $D$, the circumcenter of $\triangle DMN$ lies on the perpendicular bisector of $BC$. Therefore $BC$ is tangent to $(MDN)$. Now we finish by angle chasing: $$\angle MNY+\angle YCD=\angle MNY+\angle YND=\angle MND=\angle KDM=\angle MYB$$Hence the two circles are tangent to each other.

karitoshi wrote: Any inversion solution for this ?? Here's a solution using my favourite inversion + trig bash Let $H$ be the orthocenter of $\Delta ABC$, $L$ be the foot of perpendicular from $A$ to $BC$, $M$ be the midpoint of $XY$, $I$ be the center of $\Gamma$, and $(D)$ be the circle centered at $D$ passing through $B,C$. Apply a $\sqrt{bc}$-inversion on $\Delta ABC$ (call it $f$). Then, $f(B)=C,f(C)=B$. Let $f(X)=X',f(Y)=Y'$; then $BY//Y'C$ and $CX//X'B$, and $f(\Omega)=X'Y'$. Observe that $AI$ is the A-symmedian of $\Delta ABC$ since $I$ is the intersection of the tangents to $(ABC)$ at $B$ and $C$, so $AI$ and $AD$ are isogonal in $\angle A$. Hence, $f(\Gamma)$ is a circle with center on $AD$ passing through $B$ and $C$, i.e. $f(\Gamma)=(D)$. Therefore, the problem is equivalent to proving $X'Y'$ is tangent to $(D)$. If we let $K$ to be the projection of $D$ onto $X'Y'$, this is equivalent to showing $DK=DB$. Firstly, I claim that $\angle XDY=90^{\circ}$. This is equivalent to showing $\Delta XPD \sim \Delta DQY$ \begin{align*} &\Leftrightarrow XP\cdot YQ=DP \cdot DQ \\ &\Leftrightarrow PB^2\tan{B}\tan{C}=DP^2 \\ &\Leftrightarrow (\frac{AE}{EH})^2=(\frac{DF}{FO})^2=(\frac{DP}{PB})^2=\tan{B}\tan{C} \\ &\Leftrightarrow \frac{AL-\sqrt{LB\cdot LC}}{\sqrt{LB\cdot LC}-\frac{LB\cdot LC}{LA}}=\frac{AL-EL}{EL-LH}=\sqrt{\tan{B}\tan{C}} \\ &\Leftrightarrow \frac{AL-\sqrt{\frac{AL^2}{\tan{B}\tan{C}}}}{\sqrt{\frac{AL^2}{\tan{B}\tan{C}}}-\frac{AL}{\tan{B}\tan{C}}}=\sqrt{\tan{B}\tan{C}} \end{align*}which is true. Therefore, $M$ is the circumcenter of $\Delta XDY \Rightarrow MX=MY=MD$. Now, let $XY\cap BC=S,X'Y'\cap BC=R$. I claim that $XMYSD\sim CDBRK$. Firstly, by Menalaus' on $\Delta ABC$ and $\Delta AXY$ respectively, \begin{align*} \frac{AY'}{Y'B}\cdot\frac{BR}{RC}\cdot\frac{CX'}{X'A}=1 &\Rightarrow \frac{BR}{RC}=\frac{AX'}{CX'}\cdot\frac{BY'}{AY'}=\frac{AB}{BX}\cdot\frac{CY}{AC} \\ \frac{YS}{SX}\cdot\frac{XB}{BA}\cdot\frac{AC}{CY}=1 &\Rightarrow \frac{YS}{SX}=\frac{AB}{BX}\cdot\frac{CY}{AC}=\frac{BR}{RC} \end{align*}Together with the relative positions of the points ($S,R$ are outside segments $XY,X'Y'$), as well as the fact that $M,D$ are the midpoints of $XY,BC$, this gives $XMYS\sim CDBR$. Now, since $XY//X'Y'$, we have $\Delta MSD\sim \Delta DRK$, so we must have $XMYSD\sim CDBRK$. Notice that this implies $DK=DB$, since $MD=MY$ as proven above. Hence we are done. $\blacksquare$

Such a great problem-- but solved with a bit of nudging and prodding from another person, after a 48hr+ ordeal.

We define a load of new points: $A'=2O-A$; $E_1,E_2$ be the choices of $E$ with $AE/EH>0$ and $AE/EH<0$ respectively. We will only consider $M_1,N_1$, because the negative case is identically handled; $T_1,T_2=\overline{AO}\cap\omega$, $X=\overline{AO}\cap\overline{BC}$, corresponding to $E_1,E_2$ from earlier; $Y=\overline{T_1T_1}\cap\overline{T_2T_2}\cap\overline{BC}$ (which exists since $(BC;T_1T_2)=-1$); $Q$ as the harmonic conjugate of $A'$ wrt $BC$. Claim 1: $Q$ is the Miquel point of $ABCDUV$. (attribution needed) Proof. First, note that if $H_a$ is the reflection of $H$ in $\overline{BC}$, then $H_a,D,Q$ collinear because \[-1=(D\infty_{BC};BC)\overset{H_a}=(\overline{H_aD}\cap(ABC),A'; B,C)\]whence $Q\in\overline{H_aD}$. Now the result follows by Reim: $AH_aQC$ cyclic and $\overline{DV}\parallel\overline{AH_a}\Rightarrow Q\in(DVC)$.$\qquad\qquad\square$ Claim 2: $(AQT_1)$ touches $\omega,\overline{YT_1}$ at $T_1$. Proof. Sufficient to show $Q\in\overline{AY}$, so that the claim will follow by PoP @Y. Indeed, $-1=(BC;XY)\overset{A}=(B,C;A',\overline{AY}\cap(ABC))$ so we are done.$\qquad\qquad\square$ Claim 3: $AE_1/E_1H=AT_1/T_1A'.$ Proof. This is neither routine nor trivial (as previously thought). Construct $P=\frac{E_1+E_2}2,Z=\frac{T_1+T_2}2$ as the foot of $A$ onto $\overline{BC}$ and the foot of $O'$ onto $\overline{AO}$ respectively ($O'$ is the center of $\omega$, and the pole of $\overline{BC}$ wrt $(ABC)$). It's not too hard to show via midpoints of harmonics/etc that $Z$ is the inverse of $X$ wrt $(ABC)$. Define $B',C'=\overline{A'B}\cap\overline{AC},\overline{A'C}\cap\overline{AB}$, so that $\overline{ZB'C'}$ is the polar of $X$ wrt $(ABC)$ by Brokard. It's also easy to see that $Z,A'$ are the orthocenter and foot of $A$-altitude in $\triangle AB'C'$ (because the polar of $X\perp\overline{AOA'}$ by definition). But at the same time, observe that $\triangle ABC\overset-\sim\triangle AB'C'$! Thus the mentioned points correspond in their respective triangles which completes the claim.$\qquad\qquad\square$ We're actually finished now! $M_1,N_1\in(AQT_1)$ follows by spiral sim so we win.$\qquad\qquad\blacksquare$

one of the problems of all time [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=fuchsia; pen tri=purple; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,H,K,D,X,Y,Z,U,V,R1,R,A1,Q,CX,K1,A2,M,N,E; O=(0,0); A=dir(110); B=dir(210); C=dir(330); H=orthocenter(A,B,C); D=(B+C)/2; X=foot(A,B,C); Y=foot(B,A,C); Z=foot(C,A,B); U=extension(D,O,A,B); V=extension(D,O,A,C); CX=circumcenter(A,U,V); R1=foot(A,O,CX); R=2*R1-A; A1=O-A; Q=extension(R,A1,O,D*100); K=intersectionpoint(Q--(Q+(A-Q)*0.75),circumcircle(A,B,C)); A2=intersectionpoint(A--A1,circle(Q,distance(Q,B))); M=intersectionpoint(B--(B+(A-B)*0.9),circumcircle(A,R,A2)); N=intersectionpoint(C--(C+(A-C)*0.9),circumcircle(A,R,A2)); E=intersectionpoint(A--X,circle(D,distance(D,B))); draw(A--B--C--A,tri); draw(A--X,tri); draw(B--Y,tri); draw(C--Z,tri); draw(Q--R,pri); draw(Q--B,pri); draw(Q--C,pri); draw(Q--A,pri); draw(A--A1,pri); draw(A--U,tri); draw(U--Q,tri); filldraw(circumcircle(A,B,C),fil,pri); filldraw(circumcircle(A,U,V),sfil,sec); filldraw(circle(D,distance(D,B)),tfil,tri); filldraw(circle(Q,distance(Q,B)),sfil,sec); filldraw(circumcircle(A,R,A2),sfil,sec); filldraw(circumcircle(A,Y,Z),tfil,tri); path clip[]; clip=circle((0,0),2.5); clip(currentpicture,clip); label("$A$",A,dir(150)); label("$B$",B,dir(210)); label("$C$",C,dir(300)); label("$H$",H,dir(270)); label("$K$",K,dir(240)); label("$D$",D,dir(270)); label("$X$",X,dir(270)); label("$Y$",Y,dir(0)); label("$Z$",Z,dir(270)); label("$U$",U,dir(120)); label("$V$",V,dir(60)); label("$R$",R,dir(0)); label("$A_1$",A1,dir(300)); label("$Q$",Q,dir(270)); label("$A_2$",A2,dir(270)); label("$M$",M,dir(160)); label("$N$",N,dir(160)); label("$E$",E,dir(160)); [/asy][/asy] $U, V = OD \cap AB, AC$ $Y, Z$ = feet of altitudes from $B, C$ $H$ = orthocenter $R$ = Miquel Point of $ABCDUV$ $Q$ = intersection of tangents to $(ABC)$ at $B, C$ $AE : EH = UM : MB = VN : NC$, and $\frac{UB}{VC} = \frac{CY}{BZ} = \frac{RB}{RC} \implies R$ is the reflection of the $A$-Orthocenter Miquel point over the perp bisector of $BC$. Also note that if $A_1$ = antipode of $A$ in $(ABC)$ then $RCA'B$ is harmonic, and $\triangle{BA'C} \sim \triangle{UAV} \implies$ there is a spiral similarity at $R$ taking $BC \rightarrow UV$ and $A_1 \rightarrow A$. Define $A_2 \in AA'$ such that $AA_2 : A_2A = AE : EH$. We claim that $A_2$ is the tangency point. Note that $AMA_2NR$ is cyclic by spiral similarity. Furthermore, the homothety + reflection about $A$-bisector taking $\triangle{ABC} \rightarrow \triangle{AYZ}$ also takes $\omega \rightarrow (BC)$, $AA_1 \rightarrow AH$, $A_2 \rightarrow E \implies A_2 \in \omega$. Thus, the two circles intersect at $A_2$, and it only remains to show that they are tangent. Inverting about $\omega$, $(ABC)$ maps to itself so $R^* = A_1$ and $A^* = K = QA \cap (ABC)$, so it suffices to show that $(A_2KA_1)$ is tangent to $\omega$. $\triangle{QA_2K} \sim \triangle{QAA_2} \implies \angle{QA_2K} = \angle{QAA_1} = 90^{\circ} - \angle{A_2A_1K} \implies A_2Q$ passes through the center of $(A_2A_1K)$, thus they are tangent and we are done. $\square$

buratinogigle wrote: But this problem is mine , please the link http://forumgeom.fau.edu/FG2014volume14/FG201425.pdf v4913 wrote: one of the problems of all time Thank you for the nice comments. I also see a lot of development from this configuration. Wait for reading my upcoming paper.

I feel like this is the most natural solution. Only a sketch this time, but with motivation: Let $H$ be the orthocenter of $\triangle ABC$ (this is nice because of the Euler line). We want to remove points from the diagram, so we find $\frac{AE}{EH}=\frac{DF}{FO}=\frac{DK}{KB}=\frac{DL}{LC}$. This removes $F$ and $O$ from the diagram. Let $(BDM)$ and $(CDN)$ intersect again at $P$. We claim that $P$ is the desired tangency point (a good diagram helps here). We use the standard three steps to prove that $(AMN)$ and $\omega$ are tangent at $P$. Step 1: $P$ lies on $(AMN)$. This is just Miquel.

). It's quite clear (by say, the Humpty point or the orthocenter Miquel point) that $H$ lies on the polar of $A$ with respect to $(BC)$, so $(AH;EE')=-1 \Rightarrow \tfrac{AE}{EH}=\tfrac{AE'}{E'H}$. This means we can rewrite $(\tfrac{AE}{EH})^2$ as $\tfrac{AE}{EH} \cdot \tfrac{AE'}{E'H}=\tfrac{\operatorname{Pow}(A,\omega)}{\operatorname{Pow}(H,\omega)}$, which can be easily calculated. Step 3: $(AMN)$ and $\omega$ are tangent at $P$. This converts to $\angle BPM=\angle MAP+\angle BCP$. Now that we have $\angle MDN=90^\circ$, we can easily find $\triangle MKD \sim \triangle MDN \sim \triangle DLN$ and this combined with direct angle chasing finishes the problem.

Let $H$, $H_{a}$ be the orthocenter and foot of $A$-altitude in $\triangle ABC$ respectively. Upon drawing a good diagram one can spot the following claim. Claim. $DM\perp DN$. Proof: The claim is equivalent to $\angle MDK+\angle NDL=90^{\circ}\iff \triangle MKD\sim\triangle DLN \iff \frac{MK}{KD}=\frac{DL}{LN}$. However, note that: \[\frac{MK}{KD}\cdot\frac{LN}{DL} = \frac{BK\tan \beta}{KD}\cdot \frac{CL\tan \gamma}{DL} = \tan \beta \tan \gamma \frac{BK}{KD}\cdot \frac{CL}{DL} = \tan \beta \tan \gamma \left(\frac{OF}{FD}\right)^2 = \tan \beta \tan \gamma \left(\frac{EH}{AE}\right)^2\]\[\frac{AE}{HE} = \frac{h_{a}-EH_{a}}{AH-(h_{a}-EH_{a})} = \frac{h_{a}-\sqrt{BH_{a}\cdot CH_{a}}}{AH-h_{a}+\sqrt{BH_{a}\cdot CH_{a}}} = \frac{2R\sin \beta \sin \gamma-\sqrt{4R^2\sin \beta\sin \gamma \cos \beta \cos \gamma}}{2R\cos\alpha-2R\sin \beta \sin \gamma+\sqrt{4R^2\sin \beta\sin \gamma \cos \beta \cos \gamma}} =\]\[= \frac{\sin \beta \sin \gamma-\sqrt{\sin \beta\sin \gamma \cos \beta \cos \gamma}}{\cos\alpha-\sin \beta \sin \gamma+\sqrt{\sin \beta\sin \gamma \cos \beta \cos \gamma}} = \frac{\sin \beta \sin \gamma-\sqrt{\sin \beta\sin \gamma \cos \beta \cos \gamma}}{-\cos\beta\cos\gamma+\sqrt{\sin \beta\sin \gamma \cos \beta \cos \gamma}} = \frac{\sqrt{\sin \beta \sin \gamma}\cdot \left(\sqrt{\sin \beta \sin \gamma}-\sqrt{\cos \beta \cos \gamma}\right)}{\sqrt{\cos\beta\cos\gamma}\cdot\left(-\sqrt{\cos\beta\cos\gamma}+\sqrt{\sin \beta\sin \gamma}\right)} = \sqrt{\tan \beta \tan \gamma}\]\[\Longrightarrow \frac{MK}{KD}\cdot\frac{LN}{DL} = \tan \beta \tan \gamma \left(\frac{EH}{AE}\right)^2 = 1 \Longrightarrow DM\perp DN\]Let $T$ be the Miquel point for $D\in \overline{BC}$, $N\in\overline{CA}$, $M\in\overline{AB}$, so $T$ is the intersection of $(AMN)$, $(BDM)$ and $(CND)$. We'll show $\omega$ and $(AMN)$ are tangent at $T$. Indeed, the result from the claim implies that: \[\angle BTC = 2\pi - \angle BTM - \angle MTN - \angle CTN = 2\pi - \angle BDM - (\pi-\alpha) -\angle CDN = 90^{\circ}+\alpha\Longrightarrow T\in \omega\]So $T$ lies on both $\omega$ and $(ATN)$, so we just have to prove the two circles are tangent at $T$. This is equivalent to \[\angle BTC+\angle TMN = \angle CTN \iff \angle TMD+\angle TMN = \angle CDN \iff \angle DMN = \angle CDN \iff \triangle DMN\sim\triangle LDN\]But $\frac{DM}{DN} = \frac{MK}{DL} = MK \cdot \frac{DO}{CD\cdot DF} = \frac{MK}{DK}\Longrightarrow \triangle MDN\sim\triangle MKD\sim\triangle DLN$ and we're done.

WOW cool problem .... Lemma : Let $K,L$ be points on $BC$ such that $BK = LC$. If $M$ on $AB$ such that $MK \perp BC$ and similarly $N$ on $AC$ such that $NL \perp BC$. If $Y = (AMN) \cap (ABC)$ then $Y$ is fixed point. Proof : Let $\omega$ intersects $AB$ and $AC$ at $B'$ and $C'$ other then $B,C$. and $U = OD \cap AB$ and $V = OD \cap AC$. Redefine $Y = (AUV) \cap (ABC)$. My spiral similarity $\triangle YBU \sim \triangle YCV$. Note that $$\frac{YB}{YC}=\frac{BU}{CV}=\frac{\frac{\frac{a}{2}}{cos B}} {\frac{\frac{a}{2}}{cos C}} = \frac{cos C}{cos B}$$Using $BK = LC$ $$\frac{YB}{YC}=\frac{\frac{BK}{cos B}}{\frac{CL}{cos C}} = \frac{cos C}{cos B}$$Hence $\triangle YBC \sim \triangle YMN$. Which give us $Y \in (AMN)$ by angle simple chase. Now consider $M,N$ as given in problem. Claim : $Y \in (AB'C')$ It's well known that $B'-T-C'$ and $B',C' \in \omega$ Now note that $\measuredangle CB'C = 90 - \measuredangle CBA$ so $CB' = B'C'\cdot sin (90 - B)$ hence $$\frac{BC'}{CB'} = \frac{sin (90 -C)}{sin (90 - B)}=\frac{cos C}{cos B}$$Hence we have $\triangle YBC \sim \triangle YB'C'$ which give us $Y \in (AB'C')$. Let $R = BC \cap B'C'$ as $Y$ is Miquel point pf $\Box BCC'B'$ we have $R \in AY$. If $A' = OA \cap (ABC)$ then it's well known that $A' \in BC' , CB'$ Hence $T-A'-Y$. Consider circle $\gamma$ passing through $A,Y$ and tangent to $\omega$ at $X$. As $T$ is center $\Rightarrow \measuredangle TXR = \measuredangle TYR = 90$ By Brocard’s Theorem $AA'$ is polar of $R$ in $\omega$. As polar of $X$ pass through $R$ we get $X \in AA'$

Claim : $X \in (AMN)$ Note as $A' \in (ABC)$ if we prove $\triangle YXA' \sim \triangle YNC$ we are done. Which is basically showing $\frac{XA'}{YA'} = \frac{CN}{YC}$. Now consider $\triangle AB'C'$. $A'$ is it's orthocenter. So if $H$ is orthocenter of $ABC$, then Now using several spiral similarity $$\frac{XA'}{AA'} = \frac{EH}{AH}=\frac{OF}{OD}=\frac{LC}{DC}=\frac{CN}{CV}$$Note $\measuredangle AYV = 90 -\measuredangle B$, Which givs us the spiral similarlty which sends $UV \rightarrow BC$ also sends $A \rightarrow A'$ Hence $\triangle YAV \sim \triangle YA'C \Rightarrow \frac{AY}{YV}=\frac{YA'}{YC}$ Therefore $$\frac{XA'}{YA'}=\frac{CN \cdot AA' \cdot YV}{CV \cdot AY \cdot YC} = \frac{CN}{YC}$$

nice problem and configuration It is easy to show that $\measuredangle BEA = \measuredangle DEC$, so we construct the isogonal conjugate $T$ of $E$ with respect to $ABDC$. We claim that this is the tangency point. Note that \[ \measuredangle BTC = \measuredangle ABE + \measuredangle ECA = \measuredangle BAC + \measuredangle CEB = \measuredangle BAC + 90^\circ \]so $T$ indeed lies on $\omega$. It suffices to show that $(ANT)$ is tangent to $(BTC)$, i.e. $\measuredangle ANT + \measuredangle TCB = \measuredangle ATB$, and the other side follows similarly. Let $S$ be the $A$-antipode in $(ABC)$. We claim that $\Delta BTS \stackrel{+}\sim \Delta DNC$. Obviously $\measuredangle BST = \measuredangle DCN$. Now if $H$ is the orthocenter of $\Delta ABC$ and $\Delta XYZ$ is the orthic triangle, from $\measuredangle EZY = \measuredangle ECA = \measuredangle BCT$ and so on, we get that $AYEZH \stackrel{-}\sim ABTCS$, so \[ \frac{BS}{ST} = \frac{AS}{ST} \cos C = \frac{AH}{HE} \cos C = \frac{DO}{OF} \cos C = \frac{DC}{CL} \cos C = \frac{DC}{CN} \]which proves our claim. Now note that $T$ has an isogonal conjugate in $ABDC$, so \[ \measuredangle DTC = \measuredangle BTA = \measuredangle BTS = \measuredangle DNC \]so $D,T,N,C$ are concyclic. This finally gives \[ \measuredangle ANT + \measuredangle TCB = \measuredangle CDT + \measuredangle TCD = \measuredangle CTD = \measuredangle ATB \]as desired.

buratinogigle wrote: But this problem is mine , please the link http://forumgeom.fau.edu/FG2014volume14/FG201425.pdf I would like to re-attach this article here to prove that this problem actually originated from my research in 2014, about a year before the time of this problem.

Another one down in 30 minutes, part of the last problems of the year for me (and this one is special cause a year and some months back I thought this qn was unhuman) , It's been such a long journey and I've got 1.5 more years to show yall more cool stuff, but till then enjoy ur holidays wherever you are ) "China 2015 TST2 P3 got instant damage'd by MathLuis" (The death message is a reference to Minecraft but also how the solve went after finding a way in, you may suggest some for the future as I will be featuring them (will also give shoutouts yay)) Let $OD \cap AB=U$ and $OD \cap AC=V$ and $BH \cap AC=B_1$ and $CH \cap AB=C_1$, now from thales spam (notice that $H,G,O$ is just the euler line so it's doable and the rest is just parallel lines) and homologue points we obtain that $\triangle AC_1H \cup \{E \} \sim \triangle UDB \cup \{ M \}$ and $\triangle AB_1H \cup \{E \} \sim \triangle VDC \cup \{N \}$ (this can be found by trying to make the ratios make sense basically, and this is referred as the "instant damage pot" because it destroys the structure of the config). So now from this huge observation we have that if we let $(BMD) \cap (CND)=Q$ then from Miquel theorem we get $AMQN$ cyclic but also from the similarities $\angle BQC=\angle BMD+\angle CND=\angle B_1EH+\angle C_1EH=90+\angle BCH+\angle CBH=90+\angle BAC$ which trivially implies that $Q \in \omega$. But also to finish notice that $\angle MDN=\angle MBQ+\angle NCQ=\angle BQC-\angle BAC=90$ and also that if you let $O'$ midpoint of $MN$ then from the fact that by symetry we get $DK=DL$ by thales $O',O,D$ are colinear which now since $\angle MDN=90$ it means that $(MDN)$ is tangent to $BC$ at $D$ but now to finish we have that $\angle QBC+\angle QMN=\angle DMN=\angle NDC=\angle NQC$ which is enough to imply that $(AMN), \omega$ are tangent at $Q$ thus we are done .

Firstly, let $P$ and $Q$ be the reflections of $N$ and $M$ over $OD$, respectively. Also, let $R$ and $S$ be the intersections of $OD$ with sides $AB$ and $AC$, $X$ and $Y$ the feet of the altitudes from $C$ and $B$, respectively, and finally, let $A'$ be the reflection of $A$ over $OD$ and $H$ the orthocenter of $\triangle ABC$. Since $\omega$ is the reflection of itself over $OD$, it suffices to show that $(A'PQ)$ is tangent to $\omega$. Note that $P$ and $Q$ lie on $BS$ and $CR$, respectively, and also that \[ \frac{EA}{EH} = \frac{FD}{FO} = \frac{KD}{KB} = \frac{MR}{MB} = \frac{QR}{QC} \]and similarly, $\frac{EA}{EH} = \frac{PS}{PB}$. Moreover, note that $\triangle RAS \stackrel{+}\sim \triangle CHB$ because $\measuredangle ARS = \measuredangle BRD = 90 - \measuredangle DBR = 90 - \measuredangle CBA = \measuredangle HCB$, and analogously, $\measuredangle ASR = \measuredangle HBC$. Thus, by automatic similarity, there exists a common spiral similarity mapping $RC$ to $AH$ and to $SB$. Since $A' = BS \cap CR$, we have that $T = (A'BC) \cap (A'SR)$ is the center of the spiral similarity, and since $Y = BH \cap AS$, the quadrilaterals $TAYH$ and $TYSB$ are cyclic. Thus, $T$ is the $A$-queue point of $\triangle ABC$. Now, note that $(BXYC)$ maps to $\omega$ under the spiral similarity centered at $T$ that sends $XY$ to $BC$ (by simple angle chasing: if $J$ is the center of $(AXHY)$, then $\measuredangle NXC = \measuredangle NXH = -\measuredangle NHX = -\measuredangle AHX = -\measuredangle AYX = \measuredangle XBC$). Let $E'$ be the image of $E$ under this spiral similarity. Observe that the spiral similarity centered at $T$ mapping $CR$ to $HA$ and $BS$ satisfies $\frac{EA}{EH} = \frac{QR}{QC} = \frac{PS}{PB}$, so it maps $Q$ to $E$ and $P$. Hence, $T \in (A'PQ)$, and by automatic similarity, $TBHC \stackrel{+}\sim TPEQ \stackrel{+}\sim TSAR$. Claim: $E'$ lies on $(TPQ)$. To prove that, let $H'$ be the reflection of $H$ over $BC$. It is known that $H$ maps to $H'$ under the spiral similarity centered at $T$ that sends $XY$ to $BC$. Therefore, $TEE' \stackrel{+}\sim THH'$, and under the spiral similarity $TBHC \stackrel{+}\sim TPEQ$, $H'$ maps to $E'$ since the center is the same. Since $TBH'C$ is cyclic, $TPE'Q$ is also cyclic. Claim: $(TMN)$ and $\omega$ are tangent at $E'$. Finally, it suffices to show that $\measuredangle CE'Q = \measuredangle E'PQ + \measuredangle CBE'$, because under this angular condition, $(TMN)$ tangent at $E'$ must also be tangent to $\omega$. On the left-hand side: \[ \measuredangle CE'Q = \measuredangle CE'T - \measuredangle QE'T = \measuredangle CE'T - \measuredangle QPT = \measuredangle CE'T - \measuredangle CBT = \measuredangle BTE' + \measuredangle E'CB \]On the right-hand side: \[ \measuredangle E'PQ + \measuredangle CBE' = \measuredangle QPE + \measuredangle YXE = \measuredangle CBH + \measuredangle YBE = \measuredangle CBH + \measuredangle HBE = \measuredangle CBE \] Thus, it is sufficient to show $\measuredangle BTE' + \measuredangle E'CB = \measuredangle CBE$. It is known that $D$, $H$, and $T$ are collinear (midpoint, orthocenter, and queue point), so \[ \measuredangle BTE' + \measuredangle E'CB = \measuredangle XTE + \measuredangle E'CB = \measuredangle XTD + \measuredangle DTE + \measuredangle E'CB = \measuredangle XTH + \measuredangle DTE + \measuredangle E'CB =\]\[ \measuredangle BAH + \measuredangle HED + \measuredangle E'CB = (90 - \measuredangle CBA) + (90 - \measuredangle EDB) + \measuredangle E'CB = (180 - \measuredangle EDB) - \measuredangle CBA + \measuredangle E'CB = \]\[\measuredangle DBE + \measuredangle EYX + \measuredangle DBE - \measuredangle CBA = \measuredangle CBA + \measuredangle CBE - \measuredangle CBA = \measuredangle CBE \]which is what we wanted to prove. Hence, the two circles are tangent at $E'$. $\blacksquare$