First of all, note that $\angle HBA = \angle HKM = \angle NMF$ and similarly $\angle HCA = \angle MNF$, so $H$ and $F$ correspond to isogonal conjugates in $\triangle ABC$ and $\triangle AMN$, and so $F$ is the circumcenter of $\triangle AMN$. Now we use the tangency condition to note \[\angle MHN = \angle MKH + \angle HLN = 180^{\circ} - \angle MFN,\]and so $MHNF$ is cyclic. Then $FM = FN$, so $F$ is the arc midpoint. Now Fact 5 finishes as $FJ = FM = FA$.

Achillys wrote: $H$ lies inside the quadrilateral $BMNC$ I've heard from a friend of mine that without this condition, the statement does not hold.

Simple angle chasing yields $\angle FKH=\angle FLH = 90^\circ - \angle A$. As $MN\parallel BC\parallel KL$, we get $\angle MFN= 2\angle A$. So, $F$ is the circumcenter of $AMN$. Now the tangency gives $\angle MJN = 90^\circ + \frac{\angle MHN}2 = 180^\circ -\angle A$. So, $AMJN$ is cyclic. As $F$ is the circumcenter, the result follows.

Interesting question: Using ruler and compass construct triangle $ABC$ satisfying that problem's condition.

Draw a tangent line $t$ in $H$. $\angle HBM=\angle HCN= 90^{\circ} - \alpha $, so $t$ is the angle bisector of $\angle MHN$, which means that $J$ lies on $t$. $MN \parallel BC \parallel KL$, so $ \angle MNF=\angle HLN =\angle HCN=90^{\circ} - \alpha = \angle HBM = \angle HKM = \angle NMF$ $\triangle MNF$ is isosceles, $\angle MFN = 180^{\circ} - 2(90^{\circ} - \alpha) = 180^{\circ} - \angle MHN$, which means that $MHNF$ is a cyclic quadrilateral. $F$ is the midpoint of the arc $MN$, so $F$ lies on $t$ too. From the incenter-excenter lemma it follows that $FM=FN=FJ$ Finally, $F$ lies on the perpendicular bisector of $MN$ and $\angle MFN= 2\alpha =2\angle MAN$ which means that $F$ is the circumcenter of $\triangle MAN$. $FM=FN=FA$, and we are done.

Trivial... though constructing this is hard and so you need to work with a rough diagram Achillys wrote: Let $H$ be the orthocenter of the triangle $ABC$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively. Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$ and let $J$ be the incenter of triangle $MHN$. Prove that $F J = F A$. Angle chasing yields $\angle FMA=\angle BMK=\angle BHK=\angle HBC=90^o-C$, and so $\angle FMN=90^o-B-C$, which is symmetric in $B, C$ and so $FM=FN$. Also, $\angle MFN=180^o-2(B+C-90^o)=2A$ and so $F$ is the circumcenter of $\triangle AMN$. Next, $\angle BMH=C$ as $\odot(BMH)$ and $\odot(CNH)$ are tangent, and so $\angle KMH=90^o$. Similarly $\angle LNH=90^o$ giving $FMHN$ is cyclic, and combining with $FM=FN$ yields (by Fact 5) that $FJ=FM=FA$. $\blacksquare$

Beautiful problem (although can be difficult to do on the exam, cause of the "beautiful" diagram ). Anyway, here's my solution: Note that $\angle FKL=\angle MKH=\angle MBH=90^{\circ}-A=\angle NCH=\angle NLH=\angle FLK$. But $KL \parallel BC \parallel MN \Rightarrow \angle FMN=\angle FKL = \angle FLK =\angle FNM \Rightarrow FM=FN$. Also, $\angle FMN=\angle FNM=90^{\circ}-A \Rightarrow \angle FNM=2A \Rightarrow A$ lies on $\odot(F,FM)$. Thus, it suffices to show that $J$ also lies on $\odot (F,FM)$. Now, Let $\ell$ be the line that is tangent to both $\odot (BMH)$ and $\odot (CNH)$ at $H$. Then $\angle MHN=\angle (\ell,MH)+\angle (\ell,NH)=\angle MBH+\angle NCH=2(90^{\circ}-A)=180^{\circ}-2A\Rightarrow F$ lies on $\odot (MNH)$. But, $FM=FN$, so $F$ must be the midpoint of $\overarc{MN}$. Thus, By Fact 5, we get that $J$ also lies on $\odot(F,FM)$. Hence, proved.

Enjoyed this one, and thus posting my solution. Let $\ell$ be the line through $H$ tangent to $(BHM)$ and $(CHN)$. Since $\angle HBM=\angle HCN=\alpha$ (due to the orthocentre), $\ell$ is the angle bisector of $\angle MHN=2\alpha$. Thus $J$ lies on $\ell$. Also note that $MNLK$ is an isosceles trapezoid since $\angle HLN=\angle HKM=\alpha$ (due to cyclic quadrilaterals $MHBK$ and $NHCL$) and $MN\parallel KL$. So $FM=FN$. Now $\angle MJN=90^\circ+\frac{1}{2}\angle MHN=90^\circ+\alpha=180^\circ-\angle A$. Thus $AMJN$ is a cyclic quadrilateral. Also $\angle MFN=180^\circ-2\alpha=2(90^\circ-\alpha)=2\angle A$. All these combined forces $F$ to be the centre of $(AMJN)$ and thus $FA=FJ$ as desired. $\blacksquare$ Remark: Lines $MK,NL,\ell$ are the pairwise radical axes of $(BHM), (CHN), (MNLK)$ and thus $\ell$ passes through $F$.

First, $\angle FKH=\angle FLH=90-\angle A$, so $FMN$ is isosceles with base angle $90-\angle A$. Therefore, $F$ is the circumcenter of $AMN$. Also, as the circles are tangent, $\angle MHN=180-2\angle A$, so $\angle MJN=90+\frac{\angle MHN}{2}=90-\angle A$, and $J$ lies on circle $(AMN)$. $F$ is the center, so $FJ=FA$, as desired.

We know that both the above circles are tangent at or to center H of triangle. ABC. Also by mid point theorem and data given in the question we know that MN ∥BC ∥ KL. Then, using the cyclic quadrilaterals BUNK and CHNL, and the parallel line, a trivial angle chase gives ∠FMN = ∠FNM =90-½A. This implies that F is the circum center of ∆ AMN. Then, using the fact that is the invented of ∆ HMN and by a trivial angle chase, we find that ∠MJN = 180-A. So, A, M, N J are con cyclic w and the center of that circle is F. Hence FJ = FA and we are done. Please refer that attachment below for diagram.

We begin with the observation that $\angle MHN = \angle MBH + \angle NCH = 180 - 2 \angle A$. It follows that as $$\angle MFN = 180 - \angle FKL - \angle FLK = 180 - 2(90 - \angle A) = 2 \angle A$$that $FMHN$ is a cyclic quadrilateral. In addition, given that $FK = FL$ and $MN \parallel KL$ we see that $F$ is actually the midpoint of arc $\widehat{MN}$ on $(HMN)$. Furthermore, observe that as $\angle MFN = 2 \angle{A}$ and $FM = FN$ that $F$ is the circumcenter of $AMN$. It follows by the Incenter-Excenter lemma that $FJ = FM = FN = FA$ and we are done.

General problem. Let $H$ be the orthocenter of the triangle $ABC$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively. Assume that the circumcircles of triangles $BMH$ and $CNH$ meet again at $P$. The line through $H$ parallel to $BC$ intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$ and let $J$ be the incenter of triangle $PMN$. Prove that $FJ = FA$. Proof. We have angles chasing $$\angle MJN=90^\circ-\frac{\angle MPN}{2}=90^\circ-\frac{\angle MPH+\angle HPN}{2}=90^\circ-\frac{\angle MBH+\angle HCN}{2}=180^\circ-\angle BAC.$$ We deduce that $AMJN$ is cyclic quadrilateral. (1) Also, $$\angle FMN=\angle MKH=\angle MBH=\angle NCH=\angle NLH=\angle FNM,$$ we obtain $FM=FN.\quad (2)$ And $\angle MFN=180^\circ-2\angle FKL=2\angle BAC.\quad (3)$ From (2) and (3), we see that $F$ is circumcenter of triangle $AMN.\quad (4)$ From (1) and (4), this implies that $F$ is circumcenter of $(AMJN)$ or $FA=FJ$. We are done.

Let $F'$ be the circumcenter of $\triangle AMN$. Consider the homothety at $A$ with scale factor 2. It sends $M\to B$ and $N\to C$, so it also sends $F'\to O$, the circumcenter of $\triangle ABC$. Since $\triangle AMN \sim \triangle ABC$, $\angle AMF' = 90-C$ and $\angle ANF'=90-B$. But $\angle AMF = \angle KMB=\angle KHB=\angle BHC=90-C$, and similarly $\angle ANF = 90-C$, so $F'=F$. Now, $FA=FM=FN$, so by Fact 5 on $\triangle HMN$ is suffices to show that $FMHN$ is cyclic (then $F$ is the arc midpoint of $\widehat{MN}$). But $\angle MFN = 180-2\angle FMN$ and $\angle FMN = \angle FKH=\angle MBH=90-A$, so $\angle MFN=2A$. And using the circle tangency condition, $\angle MHN = \tfrac12 \widehat{MH} + \tfrac12 \widehat{NH} = \angle MBH+\angle NCH=2(90-A)$, so we are done.

From obvious reasons $\angle ABH=90^\circ-\alpha$ and $\angle ACH=90^\circ-\alpha \implies \angle ABH=\angle ACH$. Circumcircles of triangles $BMH$ and $CNH$ are tangent, $\angle MHN=\angle ABH+\angle ACH=2\angle ABH=180^\circ-2\alpha$. From the given facts, point $J$ is the incenter of triangle $MHN$, $\angle MJN=90^\circ+\frac{\angle MHN}{2}=180^\circ-\alpha$, so $AMJN$ is cyclic. $\angle FMN=\angle FKL=\angle ABH=90^\circ-\alpha$ and $\angle MFN=2\alpha$, so $F$ is the circumcenter of triangle $AMN$ and because of those two things, $FJ=FA$

We will show that in fact, $F$ is both the circumcenter of $\Delta AMN$ and the midpoint of arc $MN$ of $(MNH)$ not containing $H$, which will obviously finish the problem. Note that $\angle MKH = \angle MBH = \angle NCH = \angle NLH$, so $\Delta FMN$ and $\Delta FKL$ are isosceles and $F$ is the circumcenter of $\Delta AMN$. It also follows that the radical axis of the two circles bisects $\angle MHN$, and since by power of a point $F$ lies on the radical axis, we must have that $F$ is the midpoint of arc $MN$ of $(MNH)$ not containing $H$ as desired.

Note that $\angle FMN = \angle MKL =\angle MKH=\angle MBH=90-A$. Similarly $\angle FNM=90-A$. So $FN=FM$. Next note that $\angle MFN=2A$, which implies tha t $F$ is the circumcenter of $\triangle AMN$. Next note that $\angle MHN=\angle MBH +\angle MCH =180-\angle MFN$, so $F$ is the midpoint of arc $MN$ in $MNH$. Conclude with Fact 5

The key claim is that $MNLK$ is an isosceles trapezoid. As $MN\parallel KL$, it suffices to check that \begin{align*} \measuredangle MKL&=\measuredangle MKH=\measuredangle MBH=90^\circ-\measuredangle CAB,\\ \measuredangle KLN&=\measuredangle HLN=\measuredangle HCN=90^\circ-\measuredangle CAB. \end{align*}Note that since $F=MK\cap NL$, it lies on the perpendicular bisector of segment $MN$, so $FM=FN$. Then, we show two properties of $F$ to imply the result. First, we show that $F$ is the circumcenter of $\triangle AMN$. Observe that \[\measuredangle FMN=\measuredangle FKL=\measuredangle MKL=90^\circ-\measuredangle CAB.\]Since $F$ lies on the perpendicular bisector of segment $MN$, the result follows. Now, we show that $F$ is the arc midpoint of $MN$ wrt $(MHN)$ not including $H$. It suffices to check that $F$ lies on $(MHN)$, which follows from the observation that inversion about $F$ with radius $FH$ maps $M,H,N$ to collinear points $K,H,L$. Then, by incenter-excenter lemma, we have $FJ=FM=FN=FA$ as desired.

Note that because $(MHB)$ and $(NHC)$ are tangent, we have that $\angle{MHN} = 180-2\angle{A}$. Also note that $KL\parallel MN$, which gives that \[\angle{FMN} = \angle{FKL} = \angle{MBH} = 90-\angle{A} = \angle{FNM} \implies \angle{MFN} = 2\angle{A}\]which means $FNHM$ is cyclic. Note that $\angle{MHF} = \angle{FHN}$, implying that $F$ lies on the angle bisector of $\angle{MHN}$, meaning that by Fact 5, $FJ = FM = FN = FA$ because $F$ is the circumcenter of $\triangle{AMN}$.

before working on a solution, may i ask how accurate diagrams were drawn? did you just randomly move around the points A,B,C? Or is there a strategic way to get BMH tangent to CNH?

From $\angle AMF=\angle KMB=\angle KHB=\angle HBC=90-\angle C=90-\angle ANM$ and symmetry we see $F$ is the circumcenter of $AMN.$ Next we have $\angle MHJ=\angle NHJ$ implies $J$ is on the common tangent at $H$ by angle chasing, so $\angle MJN=90+\angle MHJ=90+\angle MBH=180-\angle A$ so $AMJN$ is cyclic finishing.

Absolutely awful and trivial problem. We use the tangency to get $\angle MHN=\angle MBH+\angle MCH=180 ^{\circ} -2\angle A$ so $\angle MJN=180 ^{\circ} -\angle A$ so $AMJN$ is cyclic. To finish, note that $\angle FMN=\angle NLH=\angle ACH=90^{\circ} - \angle A=\angle FNM$ which implies $F$ is the circumcenter of $AMN.$

Claim. $\overline{HJ}$ is the common tangent of $(BMH)$ and $(CMH)$. Proof. Suffices to show that the common tangent $\ell = \overline{HJ'}$ bisects $\angle MHN$. This follows as $\angle MHJ' = \angle MBH = \angle NCH = \angle NHJ'$. $\blacksquare$ Now a quick angle chase implies $\angle MHN +\angle MFN = 180^\circ$, hence $F$ lies on $(HMN)$. Furthermore, as $\angle FNM = \frac 12 \angle FHN$, $F$ is in fact the arc midpoint of $\widehat{MN}$. So it suffices to show by Fact 5 that $AMJN$ is cyclic, which is clear as $\angle MJN = 90^\circ + \frac 12 \angle MHN = 180^\circ - \angle A$.

Problem falls apart after you try to force the conditions in the diagram to hold. Let $\omega_1 = (BHM)$ and $\omega_2 = (CHN).$ First, if $X$ is a point on the common (internal) tangent to $\omega_1$ and $\omega_2,$ then $\measuredangle XHM = \measuredangle HBM = \measuredangle HBA = \measuredangle ACH = \measuredangle NCH = \measuredangle NHX,$ so $XH$ bisects $\angle MHN.$ Also, $KLMN$ is an isosceles trapezoid since $$\measuredangle MKL = \measuredangle MKH = \measuredangle MBH = \measuredangle ABH = \measuredangle HCA = \measuredangle HCN = \measuredangle HLN = \measuredangle KLN$$and $KL \parallel BC.$ Thus by Radical center on $(KLMN),\omega_1,\omega_2,$ we see that $F,J,H$ are collinear on the angle bisector of $\angle MHN.$ The isosceles trapezoid also implies that $FM = FN,$ so Incenter-Excenter tells us that $FJ = FM = FN.$ Thus it suffices to show that $F$ is the circumcenter of $\triangle AMN.$ Indeed, if we let $P$ and $Q$ be the feet of the altitudes from $M,N$ to $AN, AM,$ respectively, then $$\measuredangle AMF = \measuredangle BMK = \measuredangle BHK = \measuredangle HBC = \measuredangle PMN$$and similarly $\measuredangle ANF = \measuredangle QNM.$ Thus $F$ is the isogonal conjugate of the orthocenter of $\triangle AMN,$ which is just its circumcenter, as desired.

The following two observations conclude our proof. Firstly, notice $F$ is the center of $(AMN)$ as \begin{align*} 90 - \angle A &= \angle ABH = \angle MKH = \angle FMN \\ &= \angle HCA = \angle HLN = \angle MNF. \end{align*} Secondly, $J$ lies on this circle as \[\angle MJN = 90 + \frac{\angle MHN}{2} = 90 + \frac{\angle MBH + \angle HCN}{2} = 180-\angle A. \quad \blacksquare\]

We have $$\angle MKH = \angle MBH = 90^\circ - \angle A$$and $$\angle NLH = \angle NCH = 90^\circ - \angle A,$$so $\triangle FKL$ is isosceles. Since $MN \parallel BC \parallel KL$, we find that $\triangle FMN$ is also isosceles and $$\angle MFN = 180^\circ - 2 \angle FKH = 2 \angle A,$$so $F$ is the circumcenter of $\triangle AMN$. Note that $$\angle MHN = \angle MKH + \angle NLH = 180^\circ - 2 \angle A,$$so $\angle MJN = 180^\circ - \angle A$. Therefore, $J$ lies on $(AMN)$, so $FJ = FA$, as desired.

As $\angle FKL = \angle ABH = \angle ACH = \angle FLK$ we have that $\triangle FMN$ is isoscelles with $\angle FMN=90-\angle BAC$, thus $F$ is the circumcentre of $\triangle AMN$. Now it suffices to prove that $\angle MJN= \angle ABC +\angle ACB$ we can see this is the case as $\angle MHN = \angle 2\angle ABC+2\angle ACB +180$ from alt segment theorem due to the tangent condition, thus we have $AMNJ$ is cycle with $F$ as the centre which suffices.

Note that

We see that \[\angle FMA = \angle KMB = \angle KHB = \angle HBC = 90^{\circ}-\angle C\]and similar for $\angle FNA$. We see that $MN \parallel BC$, so \[\angle FMN = \angle FNM = 90^{\circ}-\angle A \implies FM = FN\]Therefore, a homothety at $A$ sending $M$ to $B$ and $N$ to $C$ will send $F$ to $O$, the circumcenter of $\Delta ABC$. Thus $F$ is the center of $(AMN)$. Also note that \[\angle MKH = \angle NLH = 90^{\circ} - \angle A \implies FK = FL \implies FK \cdot FM = FL \cdot FN\]Thus, $\overline{FH}$ is tangent to the two circles. Therefore, \[\angle MHF = \angle MBH = 90^{\circ} - \angle A \implies MHN = 180^{\circ} - 2\angle A \implies FMHN \text{ cyclic}\]Therefore, since $\angle MHF = \angle NHF$, incenter-excenter lemma on $\Delta MNH$ implies that $F$ is the center of $(AMJN)$, which finishes.

First notice by easy angle chasing that $MNLK$ is an isoceles trapezoid. This implies that $F$ lies on the radical axis of $(BHM)$ and $(CNH)$ or the common tangent at $H$ to these circles. Notice this bisects $\angle MHN$. Also $\angle MFN=2A$ so $F$ lies on the circle $(MON)$ where $O$ is center of $\Delta AMN$ and $F$ also lies on opposite side of $MN$ as $H$ so the only possibility is that $ F \equiv O$. This implies $F$ is midpoint of arc $MN$ not containing $H$ in $(MHN)$. So Fact $5$ gives us that $FJ=FM=FN$ but $FM=FA$ so we are done.

From the given circles tangency points ; $(BMH)$ $or$ $just$ $(KMH)$ $and$ $(CNH)$ $==>$ $ \angle MHN= \angle MKH + \angle NLH =180- \angle KFL=180- \angle MFN$ $ \angle MFN + \angle MHN=180$ ==> $MHFN$ $is$ $cyclic$. $ \angle MBH= \angle MKH = \angle ABH= 90- \angle A$ Analogic steps for $ \angle HCN$ We will get; $ \angle HLN = \angle HKM$ $ \angle LKF = \angle FLK$ From parallelity; $ \angle FMN = \angle FNM$ $FM=FN$ $FMN$ isosceles triangle H and F are isogonal conjugates. F is circumcenter of $(AMN)$ Also F is the midpoint of the arc $FA=FM=FN=FJ$ so , problem ends

Note that triangle $ABC$ is acute since otherwise $H$ would coincide with $A$ or lie outside of triangle $ABC$, contradicting the assumption that $H$ lies in quadrilateral $BMNC$. Now \[\angle FMN = \angle MKH = \angle MBH = 90^\circ - \angle BAC\]and \[\angle FNM = \angle NLH = \angle NCH = 90^\circ - \angle BAC\]due to $MN$ being parallel to $KL$; hence $FM = FN$. As such, \[\angle MFN = 180^\circ - \angle FMN - \angle FNM = 2 \angle BAC,\]and since $A$ and $F$ lie on the same side of $MN$, this implies that $F$ is the circumcentre of triangle $AMN$. Similarly, \begin{align*} \angle MFN &= 180^\circ - \angle FMN - \angle FNM = 180^\circ - \angle MBH - \angle NCH \\ &= 180^\circ - \angle(MH, \ell) - \angle(NH, \ell) = 180^\circ - \angle MHN \end{align*}where $\ell$ is the common tangent of $(BMH)$ and $(CNH)$, and since $F$ and $H$ lie on opposite sides of $MN$, this implies that $F$ is also the midpoint of arc $MN$ of $(MHN)$. Hence by the incentre-excentre lemma, $FJ = FM = FA$, as desired.

Note that \[\angle BHC=\angle BMH+\angle CNH.\]Eventually, this rearranges to $180^{\circ}-2\angle A=\angle MHN$. Now $\angle MJN=90^{\circ}+\frac12\angle MHN=180^{\circ}-\angle A$, so $AMJON$ is cyclic. Also, $\angle FMN=\angle MKH=\angle MBH=90^{\circ}-\angle A$, and similarly for $\angle FNM$. Therefore, $F$ is exactly the center of $(AMJON)$, so $FA=FJ$. $\blacksquare$