Let $G = JH\cap AX$, $ME\cap AC = L$, $D = AH \cap BC$. $\frac{AH}{HD} = \frac{LE}{EM} = \frac{AG}{GM}$, which leads to $GE\parallel AC$. By Reim's theorem $XEGF$ is cyclic. By Power of a point: $HG\cdot HJ = HF\cdot HE\rightarrow HJ = \frac{HF\cdot HE}{HG} = \frac{2AH\cdot BD}{AB}\cdot \frac{HE}{HG}$ Since $\triangle ABH \sim \triangle HEG$ the result $HJ = 2BD$ follows, this implies that $J$, $B$ and $AH\cap (ABC)$ collinear, and $\angle HJB = \angle HBD$ Let $ME\cap \omega = K$ $\angle JKB = 180 - \angle HGE = 180 - \angle AHB = \angle BHD = 90 - \angle HBD$ and this implies the desired result. $\square$

Let $AH$ meet $\odot(ABC)$ again at $D$. Let line through $H$ parallel to $BC$ intersect $AM$ at $K$ and $BD$ at $J'$. Let $BD$ meet $EM$ at $P$. Claim 1:$J'FEP$ is cyclic Proof: $BH=BD=BF$. $\angle DHJ'=90^{\circ}$ and $BH=BD$. So $BH=BD=BJ'$. So $BH=BD=BF=BJ'\implies FJ'DH$ is cyclic. $PE\parallel DH\implies\angle EPJ'=\angle HDJ'=180^{\circ}-\angle J'FH=180^{\circ}-\angle J'FE$. So we get $J'FEP$ is cyclic as desired. Claim 2: $J'KXP$ is cyclic Proof: $\angle BXM=\angle BXA=\angle BCA=\angle BDA=\angle BPM$. So $BMXP$ is cyclic. But $J'K\parallel BM$. So $\angle J'KX=\angle BMX=180^{\circ}-\angle BPX=180^{\circ}-\angle J'PX$. So $J'KXP$ is cyclic as desired. Claim 3: $J'EKP$ is cyclic Proof: Now $KE\parallel AC$ is well known. So $\angle J'KE=\angle HKE=\angle HKM+\angle MKE=\angle AMC+\angle MAC=180^{\circ}-\angle ACB=180^{\circ}-\angle EPJ'$. So $J'EKP$ is cyclic. Summing up all three claims we get that $J'FKEXP$ is cyclic. So we also get $J'$ is nothing but $J$ itself. So $JB\cap EM=P$ which indeed lies on $\omega$ as desired. Q.E.D.

One solution for this problem.

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Let $H_a$ be the reflection of $H$ across $BC$. It is well-known that $H_a\in (ABC)$. Let $J' = H_aB\cap HJ$. Claim: $J'H_aHF$ - cyclic. Proof: $H_aB = HB$ and $\angle BHJ' = 90^{\circ}\implies J'B = HB = H_aB$ and $BF = BH$. Thus, $J',H_a,H,F$ all lie on the circle with center $B$ and radius $BH$. Let $Y = BH_a\cap EM$. Claim: $J'EFY$ - cyclic. Proof: By Reim's theorem from the previous claim and $HH_a\| EM\implies J'EFY$ - cyclic. Claim: $BYXM$ - cyclic. Proof: $\angle BYM = 90 - \angle YBM = 90 - \angle HBC = \angle BCA = \angle BXA = \angle BXM$. Claim: $J'TXY$ - cyclic where $T = AX\cap J'H$. Proof: By Reim's theorem from the previous claim and $BM\| J'H\implies J'TXY$ - cyclic. Let the circumcircle of $J'TXY$ be $\Gamma$. Claim: $FJ'TX$ - cyclic. Proof: $\angle FJ'T =\angle FJ'H = \angle FH_aH = \angle FH_aA = \angle FXA = \angle FXT$. So, $F\in \Gamma$. From the second claim $\implies E\in \Gamma\implies J'FEX$ - cyclic. Therefore, $J' = J$ and $JB\cap EM = Y\in \Gamma$, as needed. Remark: The main difficulty of the problem was the introduction of $H_a$ and $T$.

Angel chasing problems are always cute! Solution : Let $\overline{AH}\cap\odot ABC=K\neq A$ and that $\overline{BK}\cap\overline{EM}=D^*$. Claim : $D^* \in \odot XEF$ Proof : Notice that $$\angle MXB=\angle AXB=\angle ACB=\angle AKB=\angle MD^*B$$which shows that $\{B,M,X,D^*\}$ are cyclic points. It follows that $$\angle XD^*E=\angle XD^*M=\angle XBM=\angle XBC=\angle XFC=\angle XFE$$which proves our desired claim. $\square$ Let $\overline{BD^*}\cap \odot XEF=J^*$. We will, instead, show that $\overline{HJ^*}\parallel \overline{CB}$ Now $\angle J^*KH=\angle J^*D^*E=180^\circ-\angle J^*FE$ so $\{J^*,F,H,K\}$ are cyclic points. So $$\angle FJ^*H=\angle FKH=\angle FKA=\angle FBA=\angle HBA=90^\circ-\angle A$$Consequently, we have $$\angle HJ^*D^*=\angle FJ^*D*-\angle FJ^*H=\angle CED^*-90^\circ+\angle A=\angle B+\angle A - 90^\circ=90^\circ-\angle C=\angle CAK=\angle CBK=\angle CBD^*$$which finishes the problem as this shows $J^* \equiv J$ and that $D^* \equiv D$. $\blacksquare$

Solved with @blastoor. Maybe a bit overcomplicated solution. We begin with introducing point $O$ on the line $AM$ such tha $HO \parallel BC$. Let $AH \cap BC = G$ and $CH \cap AB =N $ Claim: $OE \parallel AC$ Proof: Simple angle chasing reveals that $\triangle MEC \sim \triangle NHA$. This implies that: $$ \frac{AN}{AH} =\frac{MC}{CE} =\frac{GM}{HE} $$On another hand we have that $\frac{AH}{AG} =\frac{HO}{GM}$. Consequently we have that: $$ \frac{AN}{AH} = \frac{GM}{HE} = \frac{HO \cdot AG}{AH \cdot HE} \implies \frac{HO}{HE} = \frac{AN}{AG} $$But since we have that $AN \cdot AB = AH \cdot AG$ we conclude that $\frac{HO}{HE} =\frac{AH}{AB}$. Notet that $\angle BAH =\angle BCH =\angle OHE$, which implies that $\triangle BAH \sim \triangle EHO$. Thus we conclude that $\angle OEH = \angle ABH = \angle ACH$ and therefore $OE \parallel AC$ as desired. Now this implies that $\angle OEF =\angle ACF =\angle ACF$ and therefore quadrilateral $FOEX$ is cyclic. Assume that $AH \cap \odot(ABC) = I \neq A$ and define point $J'= IB \cap HO$. We will prove that $J=J'$. Note that: $$ \angle HJ'B = \angle CBI = \angle CFI$$and thus quadrilateral $J'FHI$ is cyclic. It is easy to see that $\angle J'FH = 180^{\circ} - \angle C$. But on another hand we have that $\angle HOM =\angle AMC $ and also $\angle MOE =\angle MAC$ and thus $\angle HOE =180^{\circ} - \angle C =\angle J'FH$. We conclude that quadrilateral $J'FOE$ is cyclic and consequently all five points $J', F, O, E, M$ lie on one circle $\omega$. This proves that $J = J'$ Now we define point $K = JI \cap EM$. Observe that $\angle BIA = \angle BKE = \angle C$ and thus $K$ lies also on $\omega $ and we are done.

The confusing part about this problem is that the midpoint condition is useless. Indeed, it is true for any point $M$ on $BC$ where $E$ is defined to be the intersection of the perpendicular to $BC$ through $M$ and the $C$-altitude

My solution.

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We will prove it for any $M$ on $BC$($E$ is on the line perpendicular to $BC$ at $M$). Let the circle centered at $B$ intersect $(ABC)$ at $D$ and the line through $H$ parallel to $BC$ at $J'$. $AH \cap BC=L$. $A,H,D$ are collinear. $J,B,D$ are collinear because $\angle JHD=90$. We claim that $J'$ is on $(EFX)$ which implies $J=J'$. $J'$ is on $(EFX) \iff \angle EXJ=180-\angle JFH(=\angle JDH=\angle C)$. Thus it is enough to prove that $\angle EXJ=\angle C$ . $\textbf{Claim:}$ $\triangle EMX\sim \triangle JBX$ Proof: $\angle JBX=180-\angle DBX=180-\angle DAX=180-\angle AME=\angle EMX$ and $\frac{BX}{BJ}=\frac{MX}{ME} \iff \frac{BX}{MX}=\frac{BD}{ME} \iff \frac{AC}{CM}=\frac{AC}{DL}\cdot \frac{LC}{ME} \iff \frac{ME}{MC}=\frac{LC}{LH}$ which is true.$\square$ From the above claim $\angle JXE=\angle BXM=\angle C \implies J=J'$ as desired. From the above claim we have also $\angle BJX=\angle MEX \implies JB$ and $EM$ meet on $\omega$.

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The problem is true if $M$ is any point on $BC$. We prove this generalization. Let $G$ be the intersection of the parallel line through $H$ with $AM$. Claim: $F,G,E,X$ are cyclic. Proof. It is sufficient to prove that $GE\parallel AC$ because Reim's theorem concludes. Notice $\angle GHE=\angle HCB=\angle HAB$. Let $K$ be the foot from $A$ to $BC$. Also notice \[ \frac{HG}{KM}=\frac{AH}{AK}, \frac{MK}{HE}=\frac{CK}{CH}=\frac{AK}{AB} \]Multiplying these two we get $\frac{HG}{HE}=\frac{AH}{AB}$. Hence $\triangle GHE\sim\triangle GAB\implies \angle GEH=\angle HBA=\angle ACH\implies GE\parallel AC$. $\square$ Claim: $J,F,H,L$ are cyclic with center $B$. Proof. Notice $\angle FJH=\angle FJG=\angle FXG=\angle FXA=\angle FLA=\angle FLH$ $\implies$ $J$, $F$, $H$ and $L$ are cyclic. Since $BL=BH=BF$ therefore $B$ is the center of that circle. $\square$ Since $\angle JHL=90^\circ$ and $B$ is the center therefore $J,B, L$ are collinear. Let $N=JB\cap BN$. Since the quadrilaterals $JFHL$ and $JFEN$ are similar and $JFHL$ is cyclic therefore $JFEN$ is also cyclic. And we are done.

musan1909 wrote: Let $G = JH\cap AX$, $ME\cap AC = L$, $D = AH \cap BC$. $\frac{AH}{HD} = \frac{LE}{EM} = \frac{AG}{GM}$, which leads to $GE\parallel AC$. By Reim's theorem $XEGF$ is cyclic. By Power of a point: $HG\cdot HJ = HF\cdot HE\rightarrow HJ = \frac{HF\cdot HE}{HG} = \frac{2AH\cdot BD}{AB}\cdot \frac{HE}{HG}$ Since $\triangle ABH \sim \triangle HEG$ the result $HJ = 2BD$ follows, this implies that $J$, $B$ and $AH\cap (ABC)$ collinear, and $\angle HJB = \angle HBD$ Let $ME\cap \omega = K$ $\angle JKB = 180 - \angle HGE = 180 - \angle AHB = \angle BHD = 90 - \angle HBD$ and this implies the desired result. $\square$ Why XEGF is cyclic by reim? Can u explain clearly?

Quote: Why XEGF is cyclic by reim? Can u explain clearly? Actually $XEGF$ is cyclic by the converse of Reim's Theorem. Do you know Reim's Theorem? It says- Two circles $\omega_1$ and $\omega_2$ intersects at $A$ and $B$. A line $\ell_1$ passing through $A$ intersects $\omega_1$ and $\omega_2$ at $P$ and $Q$ respectively. Another line $\ell_2$ passing through $B$ intersects $\omega_1$ and $\omega_2$ at $R$ and $S$ respectively. Then we have $PR\parallel QS$. The proof is simple angle chasing. The converse is also true.