This problem is easy, making an$\sqrt{ac}$- inversion , and radical center of 2 tangent circles and a secant that contains F, we obtain that B, F, E'(inverse of E), are collinear. I will post the solution in a few days.

dkshield wrote: This problem is easy, making an$\sqrt{ac}$- inversion , and radical center of 2 tangent circles and a secant that contains F, we obtain that B, F, E'(inverse of E), are collinear. I will post the solution in a few days. can u show your solution?

Very funny problem that dies to a very simple trick. Let $B'$ the reflection of $B$ over $AC$ and let $B_1$ be the B-antipode in $(ABC)$, let $AB_1 \cap B'C=E'$ and $AB' \cap B_1C=F'$. Claim 1: $HFOME$ is cyclic. Proof: We prove $\angle HFO=90$ (and its enough since $\angle HEO=90$ is analogous). $$360-\angle HFO=\angle BFO+\angle BFH=90+\angle BAC+180-\angle BAC=270 \implies \angle HFO=90$$Hence $HFOME$ is cyclic and has diameter $HO$ (note its trivial that $\angle HMO=90$) Claim 2: $BE, BF$ are isogonal w.r.t. $\angle ABC$ Proof: Make a $\sqrt{ac}$ inversion, and notice that $E \to E'$ and $F \to F'$ hold on the inversion, now clearly $BB', BB_1$ are isogonal w.r.t. $\angle ABC$ so by Isogonality Line Lemma (or DDIT) we get that $BE', BF'$ are isogonal on $\angle ABC$ but inverting back this gives $BE, BF$ isogonal w.r.t. $\angle ABC$ as desired. Finishing: By angle chasing and arcs knowladge: $$\angle FHM=\angle ABF=\angle CBE=\angle EHM \implies ME=MF \; \text{hence we are done :D}$$

Official Solution. Let the extension of $B H$ meet the circumcircle at point $D$. Prove that $E$ lies also on circles $D C O$ and $A D H$. In fact let $E^{\prime}$ be the second common point of circles $A B O$ and $D C O$. Then $\angle B E^{\prime} C=2 \pi-\angle B E^{\prime} O-$ $\angle C E^{\prime} O=B A O+\angle C D O=\pi-(\angle A O B+\angle C O D) / 2=\pi / 2$, i.e. $E^{\prime}$ coincide with $E$. Similarly $F$ lies on circles $C H D$ and $A O D$. Note now that $\angle O E H=2 \pi-\angle O E B-\angle B E H=\angle O A B+\angle B C H=$ $\angle C B H+\angle B C H=\pi / 2$, therefore $E$ lies on the circle with diameter $O H$. Similarly $F$ lies on this circle. It is clear that $M$ also lies on this circle. Prove that the reflections of lines $A F, B F, C F, D F$ about the bisectors of angles $A, B, C, D$ respectively meet at $E$. Let $P, Q, R, S$ be the projections of $F$ to $A B, B C, C D, D A$ respectively. Then $\angle P S R+\angle P Q R=\angle F S P+$ $\angle F S R+\angle F Q P+\angle F Q R=\angle F A B+\angle F R C+\angle F B A+\angle F D C=\pi$, because $\angle A F B=\angle C F D=\pi / 2$. Thus $P, Q, R, S$ are concyclic. Then the reflections $P^{\prime}, Q^{\prime}, R^{\prime}, S^{\prime}$, of $F$ about $A B, B C, C D, D A$ are also concyclic. Since, for example, $A P^{\prime}=A F=A S^{\prime}$, the perpendicular bisector to $P^{\prime} S^{\prime}$ coincide with the bisector of angle $P^{\prime} A S^{\prime}$, which is symmetric to $A F$ about the bisector of angle $A$. Hence the reflections of $A F, B F, C F, D F$ about the corresponding bisectors meet at the circumcenter $E^{\prime \prime}$ of $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$. It is easy to see that the angles $B E^{\prime \prime} C$ and $A E^{\prime \prime} D$ are right, i.e. $E^{\prime \prime}$ coincide with E. Finally we obtain that $\angle E H M=\angle E B C=\angle F C A=\angle F H M$ and since $E, F, H, M$ are concyclic, $E M=F M$

Claim $: MEHFO$ is cyclic. Proof $:$ Note that $\angle HEO = \angle HEB + \angle BEO = \angle HCB + \angle BAO = 90$ and similarly $\angle HFO = 90$ and obviously $\angle HMO = 90$. Let $BH$ meet $ABC$ at $D$. Claim $: AHED,CFHD$ are cyclic. Proof $:$ Note that $\angle ADH = \angle ADB = \angle C = \angle HEB = \angle AEB - \angle AEH = 2\angle C - \angle AEH$ so $AHED$ is cyclic. similarly $CFHD$ is cyclic. Claim $:DEOC,DOFA$ are cyclic. Proof $:$ Note that $\angle AFD = \angle AFH + \angle DFH = \angle ABD + \angle ACD = 180 - 2\angle A = \angle AOD$. we prove the other part with same approach. Now note that by Radical axis we have that $AB,HF,EO,CD$ and $AD,HE,OF,BC$ are concurrent. Let $AB,CD$ meet at $L$ and let $AD,BC$ meet at $K$. Let $LH$ meet $BC$ at $P$. Note that $(KP,BC)=-1$ so $(KL,TS)=-1$. Also we have $\angle SHT = 90$ so $SH$ and $TH$ are internal and external angle bisectors of $\angle LHK$. Now we have $\angle MHF = \angle LHS = \angle KHS = \angle EHM$ so $ME = MF$ as wanted.