We proceed in steps, by letting $CF \perp AB$. Lemma 1 If $DE$ meets the circumcircle of triangle $ABC$ again at $X,Y$ then if $MX,MY$ meet the nine point circle of $ABC$ again at $P',Q'$ then $F$ is the centre of spiral similarity sending $P'Q'$ to $XY$. Proof Indeed, $XP' \cap YQ'=M$ and so, since $(DEM) \cap (MXY)=\{M,F\}$ we see that the claim holds by the spiral similarity Lemma. Lemma 2 Lines $XY$, $P'Q'$ and $BA$ concur. Proof Indeed, let $XY \cap BA=T$ and see that $\angle FXT=\angle FP'M=\angle A-\angle B$ which means that points $F,P',T,X$ are con cyclic and thus, $P'Q'$ passes through $T$. Lemma 3 Line $P'Q'$ is the radical axis of $(AHB)$ and $(DEM)$. Proof Indeed, let us use power of a point. $TA.TB=TF.TM=TP'TQ'$ and thus, since $MA^2=MB^2=MX.MP'=MY.MQ'=MF.MT$ we get that $\angle AP'B=\angle AQ'B=180-\angle C=\angle AHB$ which completes the proof. These clearly, prove the result. P.S.- I am pretty sure inversion about $H$ with radius $\sqrt{HA.HD}$ and reflection in $H$ composed will also give a neat solution. Oh, darn, I messed the notation. Thanks for pointing out PRO 2000 (will fix it asap)

let $\phi$ the inversion with center $M$, and raduis $r=MA$. let $U,V=MH\cap (ABC)$ with $U$ in the arc $ACB$ , and denote $X'=\phi(X)$. then: $MU.MV=MA.MB=MH.MU \rightarrow \phi(H)=U$, thus we get $\phi(AHB)=(ABC)$. further more $P',E,D,Q'$ are collinear with ${P',Q' }\in (ABC)$. hence $Q'=MQ\cap DE$. now, consider $\psi$ the composition of an inversion with center $H$ and raduis $\sqrt{HA.HD}$, and reflection wrt $H$. then, $\psi(ABC)=(MED)$, and $\psi(DE)=(ABH)$ therefore $\psi(Q')=(MED)\cap (ABH)=P$ (the latter follows since $\psi(Q')$ is on the same side of  $A$ wrt $MH$. hence done.

My solution: Let $R=MH\cap \odot(ABC)$ $\Longrightarrow$ $\angle MRC=90^{\circ}$ since $ABCR$ is cyclic we get $\angle ARM=\angle HAM$ $\Longrightarrow$ $AM^2=MH.MR$. Let $X,Y=\overline {DE}\cap \odot(ABC)$ such that $X,D,E,Y$ they are in that order and let $P'$ and $Q'$ points in $MY$ and $MX$ such that $ME^2=MD^2=MQ'.MX$ and $MD^2=ME^2=MP'.MY$ $\Longrightarrow$ $\angle QEM=\angle EXM=\angle MDQ$ $\Longrightarrow$ $EDQ'M$ is cyclic similarly $EDP'M$ is cyclic $\Longrightarrow$ $P',Q'\in \odot (DEM)...(1)$. We consider the inversion $\mathbf{I}$ with center $M$ and radius $MA$ $\Longrightarrow$ $\mathbf{I}(\odot (AHB))=\odot (ARB)=\odot (ABC)...(1)$, but $MA^2=ME^2=MP'.MY$ $\Longrightarrow$ $\mathbf{I}(Y)=(P')$ since $Y\in \odot (ABC)$ $\Longrightarrow$ $P'\in (ABH)$ similarly $Q'\in (ABH)$ $\Longrightarrow$ $P',Q'\in \odot (ABH)...(2)$. By $(1)$ and $(2)$ we get $P'=P$ and $Q'=Q$. Let $O'$ the circumcenter of $\triangle ABH$ and let $N$ the circumcenter of $\triangle DEM$ $\Longrightarrow$ $C,N,O'$ are collinear, but $NO'$ is the perpendicular bisector of $PQ$ $\Longrightarrow$ $CP=CQ$. Futhermore it's well know $CY^2=CE.CA=CX^2$ and $MP.MY=MQ.MX$ $\Longrightarrow$ $XYPQ$ is cyclic. Since $C$ it belongs in the perpendicular bisectors of $XY$ and $PQ$ we get $C$ is the center of $\odot (XYPQ)$. By radical axis in $\odot (DEM)$, $\odot (AHB)$ and $\odot (ABDE)$ we get $PQ$, $AB$, $DE$ are concurrent in $S$. By Brocard's theorem in $\odot (ABDE)$ we get the orthocenter of $\triangle MCS$ is $BE\cap AD=H$. By Brocard's theorem in $\odot (XYPQ)$ we get the orthocenter of $\triangle CMS$ is $QY\cap PX$ $\Longrightarrow$ $H=QY\cap PX$ hence $PH$, $MQ$ and $DE$ are concurrent in $X$.

@ #2 Probably $CF \perp AB$. Do you mean $XY \cap BA=T$ ? (This is because $XY \cap BC=D$ ). This probably leads us to $\angle FXT=\angle FP'Q'$ by lemma 1 . Can you please explain why $\angle FXT=\angle FP'M=\angle A-\angle B$ . ( Sorry as i am dumb ) . By my def of $T$ , we get that it is radical axis of $(AHB) , (DEM)$ by converse of radical axis theorem on $(AHB) , (DEM) , (AEDB)$ , so i cant get lemma 3 ? So please explain your solution . It is very awesome !!!

Consider the inversion $\Psi$ with center $H$ fixed the circumcircle of $\triangle ABC.$ Denote by $A^*$ the image of their corresponding point and $\Gamma$ the circumcircle of $\triangle ABC.$ Let $X^*\equiv P^*H\cap \Gamma,S\equiv Q^*H\cap \Gamma.$ Now it suffices to prove that: $M^*,H,X^*,Q^*$ are concyclic; $B^*,X^*,H,E^*$ are concyclic. Which is not hard to approach, notice that $\Gamma$ and $\odot(DEF)$ are homothety with center $H$ and ratio $\tfrac{1}{2},$ so \[HD^*\times HA^*=HE^*\times HB^*=HN\times HM^*=HS\times HQ^*=HX^*\times HP^*=-2\times\mathcal{P}(H,\Gamma).\]Hence we get a number of circles. By Reim's Theorem we deduce that $N$ is the intersection of $Q^*X^*$ and $P^*S,$ this is because the tangent passing through the midpoint of $\overline{D^*E^*}$ WRT $\Gamma$ is parallel to $A^*B^*.$ The remaining job is just angle-chasing, and we're done! $\square$ [asy][asy] size(8.5cm); pathpen=black; pointpen=black; pointfontpen=fontsize(9pt); void b(){ pair F=D("F^*",dir(70),dir(70)); pair E=D("E^*",dir(-35),dir(-35)); pair D=D("D^*",dir(-145),dir(-145)); pair H=D("H",IP(foot(F,D,E)--F,D--foot(D,E,F)),dir(90)*1.5); pair N=D("N",midpoint(D--E),dir(-90)); pair M=D("M^*",IP(L(N,H,0,4),circumcircle(D,E,F)),dir(45)); pair A=D("A^*",foot(D,E,F),dir(50)); pair B=D("B^*",foot(E,D,F),dir(100)*1.5); pair P=D("P^*",IP(L(B,A,0,5),circumcircle(D,E,F)),dir(10)); pair Q=D("Q^*",IP(L(B,A,5,0),circumcircle(D,E,F)),dir(170)); pair X=D("X^*",IP(circumcircle(B,A,N),L(H,P,5,0)),dir(-135)); pair S=D("S",OP(N--P,circumcircle(N,A,B)),dir(-45)); D(circumcircle(D,E,F)); D(D--E--F--cycle); D(P--Q); D(circumcircle(B,A,N),red+linetype("4 4")); D(N--M,dashed); D(P--N); D(Q--N); D(Q--S,dashed); D(P--X); D(circumcircle(Q,M,H),magenta+linetype("4 4")); MP("\Gamma",(0.3,0.45),S); } b(); pathflag=false; b(); [/asy][/asy] In the diagram $F$ is the foot from $C$ to $\overline{AB},$ and $F^*$ is its image.

My solution: Let $ED$ cut $AB$ at S. Since $EDBA$ is concyclic, then $P,S,Q$ are collinear. Inversion with center $H$, radius $HE.HB$: $(O) \mapsto (9-points)$, $(HAB) \mapsto ED$ $\Longrightarrow$ $P \mapsto L$ ( with $L$ is the intersection of $(O)$ and $ED$$\Longrightarrow$ $ED,PH$ cut each other at $(O)$. Inversion with center $M$, radius $MA^2$: $(AHB) \mapsto (O)$, $(9-points) \mapsto ED$ $\Longrightarrow$ $Q \mapsto L$ $\Longrightarrow$ the lines $ED, PH,$ and $MQ$ all pass through a single point on the circumcircle of $\triangle ABC$

solution

$(MED)$ is the nine-point circle of $ABC$. Inversion under $I(M, AM)$, sends $H \rightarrow H'$, where $H'\in (ABC)$ (by Brokard's Theorem). $(ABH) \rightarrow (ABC)$, and $(MED)\rightarrow ED$, $Q\rightarrow Q',\ P\rightarrow P'$, where $Q', P' \in ED$, and $P', Q'\in (ABC)$. Proof to follow: $PH \cap (ABC)=Q'$. So, the composition of an inversion $I(H,\sqrt{HE\cdot HB})$ and a reflection w.r.t. $H$, sends $(MED)\rightarrow (ABC)$ and $(ABH)\rightarrow DE$, which means that $P \rightarrow DE\cap (ABC)$ which implies directly that $PH\cap (ABC) =Q'$. QED.

AlcumusGuy wrote: Let $\triangle ABC$ be an acute-angled triangle with altitudes $AD$ and $BE$ meeting at $H$. Let $M$ be the midpoint of segment $AB$, and suppose that the circumcircles of $\triangle DEM$ and $\triangle ABH$ meet at points $P$ and $Q$ with $P$ on the same side of $CH$ as $A$. Prove that the lines $ED, PH,$ and $MQ$ all pass through a single point on the circumcircle of $\triangle ABC$. Let $K = DE \cap PH$ and let $L$ be the reflection of $Q$ over $M$. Note that $M$ lies on $(ABC)$. Let $F$ be the foot of the $C$-altitude. Let $K’ = MQ \cap (ABC)$. Claim 1: $A, P, D, K$ are concyclic. Proof. $\measuredangle{HDR} = 180^{\circ} - \measuredangle{EDH} = 180^{\circ} - \measuredangle{HBA} = \measuredangle{APH} $ $\quad \square$ Similarly, $P, B, K, E$ are also concyclic. Claim 2: $K$ lies on $(ABC)$. Proof. $\measuredangle{AKB} = \measuredangle{HKB} - \measuredangle{HKA} = \measuredangle{PEH} - \measuredangle{PDH} = \measuredangle{ACB} \quad \square$ Claim 3: $K’$ lies on $ED$. Proof. Note that $MQ. MK’ = MK’.ML = MD^2$. So, $(DK’Q)$ is tangent to $MD$. Similarly, $(EK’Q)$ is aslo tangent to $ME$. So, $\measuredangle{MK’D} = \measuredangle{MDQ} = \measuredangle{MEQ} = \measuredangle{MK’E}$ $\quad \square$ This shows us that $K \equiv K’$. So, $MQ, ED, PH, (ABC)$ all concur at $K \quad \blacksquare$

Consider the inversion centered at $M$ with power $MA=MB$, where $Z^*$ refers to the image of $Z$. Also, let $DE\cap PH=X$. Claim: $H^*$ lies on $(ABC)$. Proof: Let $HM$ intersect $(ABC)$ again at $H'$. by reflecting the orthocenter, $MH'\cdot MH^*=MH\cdot MH^*=MA\cdot MB$. Claim: $P^*$ lies on $(ABC)$ and $DE$. Proof: Let $DE\cap (ABC)=P'$. Because $ME=MD$, we have $\angle MEP'=180-\angle MED=180-\angle MDE=\angle MPE$ and $\widehat{MP}=\widehat{MD}-\widehat{PE}\rightarrow PEM=\angle EP'M$, hence $\triangle PME\sim\triangle EMP'$ and $\angle P'ME=\angle EMP$, implying our claim. Now, consider the negative inversion swapping the nine-pt circle and the circumcircle centered at $H$; it swaps $(AHB)$ and $DE$, and hence also swaps $Q$ and $P^*$. It also, obviously, swaps points $M$ and $H^*$. We have $HP^*\cdot HQ=HM\cdot HH^*$, so $P^*QMH^*$ is cyclic. Because $Q$ lies on both $(MP^*H^*)$ and $(MED)$, we can conclude that $Q=X^*$, which solves both parts of the problem.

Complex2Liu wrote: Consider the inversion $\Psi$ with center $H$ fixed the circumcircle of $\triangle ABC.$ Denote by $A^*$ the image of their corresponding point and $\Gamma$ the circumcircle of $\triangle ABC.$ Let $X^*\equiv P^*H\cap \Gamma,S\equiv Q^*H\cap \Gamma.$ Now it suffices to prove that: $M^*,H,X^*,Q^*$ are concyclic; $B^*,X^*,H,E^*$ are concyclic. Which is not hard to approach, notice that $\Gamma$ and $\odot(DEF)$ are homothety with center $H$ and ratio $\tfrac{1}{2},$ so \[HD^*\times HA^*=HE^*\times HB^*=HN\times HM^*=HS\times HQ^*=HX^*\times HP^*=-2\times\mathcal{P}(H,\Gamma).\]Hence we get a number of circles. By Reim's Theorem we deduce that $N$ is the intersection of $Q^*X^*$ and $P^*S,$ this is because the tangent passing through the midpoint of $\overline{D^*E^*}$ WRT $\Gamma$ is parallel to $A^*B^*.$ The remaining job is just angle-chasing, and we're done! $\square$ [asy][asy] size(8.5cm); pathpen=black; pointpen=black; pointfontpen=fontsize(9pt); void b(){ pair F=D("F^*",dir(70),dir(70)); pair E=D("E^*",dir(-35),dir(-35)); pair D=D("D^*",dir(-145),dir(-145)); pair H=D("H",IP(foot(F,D,E)--F,D--foot(D,E,F)),dir(90)*1.5); pair N=D("N",midpoint(D--E),dir(-90)); pair M=D("M^*",IP(L(N,H,0,4),circumcircle(D,E,F)),dir(45)); pair A=D("A^*",foot(D,E,F),dir(50)); pair B=D("B^*",foot(E,D,F),dir(100)*1.5); pair P=D("P^*",IP(L(B,A,0,5),circumcircle(D,E,F)),dir(10)); pair Q=D("Q^*",IP(L(B,A,5,0),circumcircle(D,E,F)),dir(170)); pair X=D("X^*",IP(circumcircle(B,A,N),L(H,P,5,0)),dir(-135)); pair S=D("S",OP(N--P,circumcircle(N,A,B)),dir(-45)); D(circumcircle(D,E,F)); D(D--E--F--cycle); D(P--Q); D(circumcircle(B,A,N),red+linetype("4 4")); D(N--M,dashed); D(P--N); D(Q--N); D(Q--S,dashed); D(P--X); D(circumcircle(Q,M,H),magenta+linetype("4 4")); MP("\Gamma",(0.3,0.45),S); } b(); pathflag=false; b(); [/asy][/asy] In the diagram $F$ is the foot from $C$ to $\overline{AB},$ and $F^*$ is its image. hello @Complex2Liu, Can you explain why "By Reim's Theorem we deduce that $N$ is the intersection of $Q^*X^*$ and $P^*S,$ this is because the tangent passing through the midpoint of $\overline{D^*E^*}$ WRT $\Gamma$ is parallel to $A^*B^*.$ " i dont understand it.

Negative inversion about $H$ with radius $\sqrt{HA\cdot HD}$ gives, $(DEM) \longleftrightarrow (ABC)$ $\Rightarrow P \longleftrightarrow P'$ where $P'$ is a point on on $(ABC)$. Now since $A,P,H,B$ are concyclic, $A',B',P'$ are collinear and thus $D,E,P'$ are collinear. $\Rightarrow HP$ and $DE$ concurr at $(ABC)$. It remains to prove that $MQ$ passes through this point too. We invert about $(ABDE)$ instead. Note that $M$ is the centre of $(ABDE)$. This maps, $(DEM) \longleftrightarrow DE$ and $(AHB) \longleftrightarrow (ABC)$ so $Q''$ lies on both $DE$ and $(ABC)$ $\blacksquare$ $\textbf{Remark:}$ The first inversion is quite similar to IMO 2015/3.

Casey's Theorem : Suppose that $\omega_1$ and $\omega_2$ are two circles with centers $O_1$ and $O_2$ and $X$ is an arbitrary point in the plane. Then if $H$ is the foot of altitude from $X$ to the radical axis of these circles , so we have : $$|P_{\omega_1}^{X}-P_{\omega_2}^{X}|=2XH.O_{1}O_{2}$$ ( Replace $D$ with $F$ and change $A , B , C$ as $B , C , A$ and suppose that $AC > AB$ ) Name the circumcircle of the triangle $\triangle ABC$ and it's nine-point circle as $\omega$ and $\omega'$ respectively. Suppose that $EF$ intersects circle $\omega$ at $K$ which $K$ lies on the arc $\widehat {AC}$. Then consider the unique point $Q$ on the line segment $KM$ such that : $$KQ=2OM.\sin{(\angle KMC) } (I)$$Now firstly we put $Q$ as the intersection point of $KM$ and $\omega '$ which is nearer to $AC$ and then as the intersection point with circumcircle of triangle $\triangle BHC$ , and we'll show that the equation $I$ holds for both of these points and as the result , we can get $Q \in (\omega') \cap (\odot BHC)$. Now if $H_K$ is the foot of altitude from $K$ to $BC$ , so by Casey's theorem for point $K$ and circle $\omega$ and circumcircle of cyclic quadrilateral $BEFC$ , while $P_{\omega}^{K}=0$ we have : $$P_{\odot BEFC}^{K}=2KH_{K}.OM \implies P_{\omega'}^{K}=KQ.KM=KF.KE=P_{\odot BEFC}^{K}=2KH_{K}.OM \implies KQ=2OM.\sin{(\angle KMC) }$$So if $Q$ lies on the circumcircle of triangle $\triangle BHC$ , then suppose that $H_Q$ is the foot of altitude from $Q$ to $BC$ and let $QH_Q$ intersects $\omega$ at point $A'$ on the arc $\widehat {BAC}$. So while $\angle BQC=180-\angle BA'C$ , we have $Q \equiv H_{\triangle A'BC}$ and as the result $KQ$ passes trough the antipode of $A'$ in $\omega$ and $\angle QKA'=90$. So one can see that : $$\triangle KQA' \sim \triangle KMH_K \implies KQ.KM=KH_K.A'Q=2OM.KH_K \implies KQ=2OM.\sin{(\angle KMC) }$$Hence we have $Q \in (\omega') \cap (\odot {BHC})$. Now suppose that $KH$ intersects circumcircle of triangle $\triangle BHC$ at point $P$ for second time and we'll show that $P$ lies on the circle $\omega'$ , so while the center of the circle $\odot {BHC}$ is the reflection of $O$ with respect to $M$ , again by Casey's theorem for point $K$ and circles $\omega$ and $\odot {BHC}$ we have : $$P_{\odot {BHC}}^{K}=KH.KP=4KH_K.OM=2P_{\omega'}^{K}=2KF.KE (II)$$Hence if the line $KH$ intersect circle $\omega'$ at point $L$ which is nearer to $AC$ , then since $H$ is the external homothety center of $\omega'$ and $\omega$ with ratio $2$ , we can get $LH=LK$ and by $(II)$ one can see that : $$KH.KP=2KL.KP=2KF.KE$$And as the result , quadrilateral $FEPL$ is cyclic and $P$ lies on the circle $\omega'$. So we're done.

Consider the inversion centered at $C$ swapping $(E,A)$ and $(D,B)$. Note that the intersections $X$ and $Y$ of $DE$ with $(ABC)$ (with $X$ and $A$ on the same side of $CH$) are fixed by this inversion. Now let $F$ be the foot from $C$ to $AB$. Note that $Y' = PH\cap (CPF)$ is also fixed by the inversion. But since the reflections of $H$ over $F$ and $P$ lie on $(ABC)$, $Y'$ also lies on $(ABC)$ by PoP, so $Y' = Y$. Similarly, we get $X = HQ\cap (CQF)$. Then Brocard on $XPQY$ (which is cyclic centered at $C$) tells us that $XP\cap YQ$ is the orthocenter of $C$, $H$, and $DE\cap AB$, which is $M$, as desired.

What a great problem. We proceed by two inversions. First, invert about $M$ with radius $MA$. Notice: Let $H'$ be the reflection of $H$ over $M$, which lies on $(ABC)$. Then, $H$ is sent to the other $\overline{MH} \cap (ABC)$ under the inversion. $A$ and $B$ are fixed, so $(ABH)$ goes to $(ABC)$; $D$ and $E$ are fixed, so $(DEM)$ goes to $\overline{DE}$. Thus $P$ and $Q$ go to the intersections of $\overline{DE}$ with $(ABC)$. In particular, for $X = \overline{PQ} \cap (ABC)$ on minor arc $\widehat{BC}$, $X$ lies on $\overline{MQ}$. Next, (negatively) invert about $H$ with radius $\sqrt{HA \cdot HD}$. Notice: $A$ is swapped with $D$ and $B$ with $E$, so $(ABH)$ goes to $\overline{DE}$; $(DEF)$ goes to $(ABC)$. Thus $P$ and $Q$ again go to the intersections of $\overline{DE}$ with $(ABC)$. Thus, $X$ lies on $\overline{HP}$ too, and we are done.

We will split this problem into two parts. For the first part, we will show that $MQ$ and $DE$ intersect on $(ABC)$. To do this, invert around $M$ with radius $MA=MB$. In this case, note that by the harmonic lemma, we have $(AB;FF*)=-1$, thus $F*$ lies on $DE$. Note that $D$ and $E$ do not change. Furthermore, the nine-point circle becomes line $DE$, and $(ABH)$ becomes $(ABC)$, so $P$ and $Q$ are now the intersections of $DE$ with $(ABC)$. Thus, $Q*$ lies on $DE$ and $(ABC)$, and by definition of inversion also lines on $MQ$, finishing this part. For the second part, we will show that $PH$ and $DE$ intersect on $(ABC)$. Invert around $H$ (negatively) such that the circumcircle and nine-point circle are swapped. Then, let $N$ be the intersection of $DE$ and $(ABC)$. We will show that $N$ and $P$ swap under this inversion, which will finish the problem. Since $D,E$ map to $A,B$, line $DE$ becomes $(ABH)$. Furthermore, $(ABC)$ becomes the nine-point circle. Hence, the image of $N$ is an intersection between $(ABH)$ and the nin-point circle on the other side, also known as $P$, hence done.

Let $R, S$ be intersections of $ED $ and $(ABC)$. Suppose $R$ lies on the arc $ABC$ and $S$ lies on the arc $BAC$. Let $F$ be foot of altitude $C$ from $AB$. Claim: $P, H, R$ are collinear and $Q, H, S$ are collinear. Proof: Consider negative inversion centered $H$ with radius $\sqrt{AH \cdot HD}$. Then this inversion swaps $(DEF)$ to $(ABC)$ and swaps $(ABH)$ to $DE$. Thus $P^*, Q^*$ are intersections of $DE$ and $(ABC)$. Therefore $R = P^*$ and $S = Q^*$.$\blacksquare$ Thus $PH \cdot HR = QH \cdot HS$, hence $PQRS$ is cyclic. Claim: $C$ is center of $PQRS$. Proof: Perform inversion centered $H$ with radius $\sqrt{CH \cdot CF}$. Then this inversion swaps $(DEF)$ to $(ABH)$, so $P, Q$ are fixed under the inversion. Note that this inversion swaps $(ABC)$ to $DE$, so $R, S$ are fixed under the inversion. Thus $CP = CQ = CR = CS = \sqrt{CH \cdot CF}$.$\blacksquare$ Now we'll prove that $M = RQ \cap PS$. Let $M' RQ \cap PS$. Note that the the three radical axis of circles $(ABH), (ABC), (PQRS)$ are concurrent. Thus let $T = AB \cap DE \cap PQ$. Let $X$ be Miquel point of quadrilateral $ABDE$. Then it's well known that $X$ lies on $CT$ and $MH \perp CT$. And note that $CH \perp MT$. This forces $H$ is orthocenter of triangle $CMT$. Thus $C$ is orthocenter of triangle $MHT$. On the other hand, by Brocard's theorem $C$ is orthocenter of $THM'$. This forces $M = RQ \cap PS$. Therefore $ED, MQ, PH$ are concurrent on $R$ and $R$ lies on $(ABC)$. This completes proof. $\blacksquare$

Solved with GrantStar First, invert about $M$ with radius $AM.$ We can check that this sends $(DEM)$ to $DE$ and $(ABH)$ to $(ABC),$ so $P,Q$ get sent to the intersections of $DE$ with $(ABC).$ Next, take a negative inversion at $H$ with radius $\sqrt{AH \cdot DH}.$ This sends $(DEM)$ to $(ABC)$ and $(ABH)$ to $DE,$ so this also sends $P,Q$ to the intersections of $DE$ with $(ABC).$ Letting $DE \cap MQ=K,$ we get that either $PH$ or $QH$ passes through $K.$ However, $QH$ cannot, so $PH$ does, and we are done.

Let the $C$-Queue point be $X$ and $DE$ intersect $(ABC)$ at $K$ and $L$, with $K$ is closer is $A$. Inverting about $H$ with radius $\sqrt{HA \cdot HD}$ composed with a reflection about $H$, we know $P \mapsto L$ and $Q \mapsto K$, so $PHL$ and $QHK$ collinear. $ED \mapsto (HAB)$, $PH \mapsto PH$, and $MQ \mapsto (HXK)$. Hence all we need to show is that $HXKP$ is cyclic, which holds as $P \mapsto K$ and $H \mapsto X$ under inversion about $(ABDE)$. $\blacksquare$

We do lots of inversion (2 times), first $A-center$ everything Let $K,L = EF \cap (ABC)$ By $\sqrt{HA \cdot HD}$ inversion we get $K-H-Q$ By $MB$ inversion we get $K-P-M$ Also note that by $\sqrt{AH \cdot AD}$ inversion, we get $AP=AK=AL=AQ$

Attachments:

Note $ME=MF$ are tangent to $(CDE)$ Now take inversion at $M$ with radius $ME$ as $(CDE)$ will remain orthogonal we get $$H \rightarrow L$$where $L = (CDE) \cap (ABC)$ hence $$ (AHB) \rightarrow (ABC)$$from $$(DEM) \rightarrow DE$$we get $Q'$ lie on $MQ,DE$ and $(ABC)$ Now take $HA.HD$ inversion swaps ninepoint circle to $(ABC)$ $$(HAB) \rightarrow (ABQ')$$ hence $P'$ lie on $(ABC)$ and $DE$ which is $Q'$ only

Subjective Rating (MOHs)

$ $

Attachments: