Let the midpoints of sides $FB$ and $FC$ be $X$ and $Y$ I will prove that $X \equiv P$ and $Y \equiv Q$. If this is proved we are done. First we observe that $NH=NF.(E,H ,F)$ lie on a circle centered at $N$) and $YF=YH$ for the same reason.$\Rightarrow \angle YHN=\angle YFN$. Now let $M'$ be the reflection of $F$ over $M \Rightarrow FBM'C$ is a parallelogram. Observe that $\frac{AE}{EC}=\frac{AD}{BC}= \frac{AF}{FB}=\frac{AF}{CM'} \Rightarrow \frac{AE}{AF}=\frac{EC}{CM'}$ and $\angle M'CE= \angle BDE (BD||M'C)=\angle EAF \Rightarrow \Delta EAF \sim \Delta ECM' \Rightarrow \angle AFE= \angle EM'C.$ Now we make the important observation that the two figures $NXMY$ and $EBM'C$ are homothetic centering $F$ mapping $M$ to $M' ,X$ to $B ,Y$ to $C$ and $N$ to $E.\Rightarrow \angle NMY = \angle EM'C= \angle EFA$..Hence $\angle NHY+\angle NMY=\angle NFC+\angle EFA = 180$ proving that $N,M,Y,H$ are concyclic .Done!

what do they mean by meet only at p and q? like its tangent? or meet the segment

liuyj8526 wrote: what do they mean by meet only at p and q? like its tangent? or meet the segment meet at segment

Actually, it's very simple if you can see that $ MG $ = $ MH $, $ NG $ = $ NH $ (Stewart's theorem), $ \angle NMG $ = $ \angle NMH $, triangle $ NMG $ is congruent to triangle $ NMH $ and $ NQ \perp FH $. Then $ NQ $ || $ EC $ and Q is the midpoint of $ FC $. Similarly, on the other side $ NP $ || $ EB $ and P is the midpoint of $ FB $. Thus $ PQ $ || $ BC $.

Let $X,Y$ be the midpoints of $BF,BE$, and $w$ the circumcircle of $XNY$, $Z$ the intersection of $w$ with $AB$. Then $\angle XNY =\angle FBE =\angle FCE =\angle XMY$, so $M$ lies on $w$ as well. $XN \parallel ZY$, so $NZ=XY=FN$, which means $Z=G$, so $P=X$.

Vo Duc Dien wrote: Actually, it's very simple if you can see that $ MG $ = $ MH $, $ NG $ = $ NH $ (Stewart's theorem), $ \angle NMG $ = $ \angle NMH $, triangle $ NMG $ is congruent to triangle $ NMH $ and $ NQ \perp FH $. Then $ NQ $ || $ EC $ and Q is the midpoint of $ FC $. Similarly, on the other side $ NP $ || $ EB $ and P is the midpoint of $ FB $. Thus $ PQ $ || $ BC $. I don't think $MG=MH$ You see here

Let $P',Q'$ be the midpoints of $BF,CF$ respectively.Let $\angle{FAB}=\angle{FDC}=\alpha,\angle{FBA}=\angle{FCD}=\theta$.Then note that $\frac{HQ'}{Q'M}=\frac{CF}{BF}=\frac{CD}{AB}$ and $\frac{HF}{FG}=\frac{FC}{FB}=\frac{CD}{AB}$.Also note that $\angle{HFG}=\angle{DFA}+\angle{DFH}+\angle{AFH}=90-\alpha+90-\alpha+\theta+\alpha=180+\theta-\alpha$ and $\angle{HQ'M}=\angle{HQ'F}+\angle{FQ'M}=2\theta+180-\theta-\alpha=180+\theta-\alpha$.Thus $\triangle{HQ'M} \sim \triangle{HFG}$.Let $\angle{GHF}=\angle{MHQ'}=x$.Since points $H,F,G,E$ are concyclic we have $\angle{FEG}=x$ and $\angle{FEH}=\angle{FGH}=\angle{HMQ'}=\alpha-\theta-x$.Thus since $N$ is the center of $\odot{HFGE}$ we have $\angle{NHF}=90-\alpha+\theta+x$.Also note that $\angle{FHM}=90-\angle{MHQ'}-\angle{Q'HC}=90-\angle{MHQ'}-\angle{Q'CH}=90-x-\theta$.Thus $\angle{NHM}=\angle{NHF}+\angle{FHM}=90-\alpha+\theta+x+90-x-\theta=180-\alpha$.Also note that $\angle{NQ'M}=\angle{NQ'F}+\angle{FQ'M}=\angle{FCE}+\angle{AFB}=\theta+180-\alpha-\theta=180-\alpha$.Thus $NHQ'M$ are concyclic.Similarly using the fact that $\triangle{MP'G} \sim \triangle{HFG}$ we can prove that $NGP'M$ are concyclic.Thus $P' \equiv P,Q' \equiv Q$ and the result is immediate.

I will prove that the circumcircle of $ \triangle{MNH} $ passes through the midpoint of segment $ FC $ (which will also demonstrate that the circumcircle of $ \triangle{MNG} $ passes throught the midpoint of segment $ FB $ which will immediately imply the desired result). Let $ X $ be the unique point such that quadrilateral $ FBXC $ is a parallelogram and let $ Y $ be the reflection of $ F $ over $ CD. $ Considering the homothey centered at $ F $ with ratio $ 2, $ it clearly suffices to show that points $ E, X, C, Y $ are concyclic. Note that $ \angle{XCY} = 360 - \angle{YCF} - \angle{FCB} - \angle{BCX} = 360 - 2\angle{DCA} - \angle{ACB} - \angle{CBD} = 360 - \overarc{BC} - \frac{\overarc{AB}}{2} - \frac{\overarc{CD}}{2} = 180 - \left(\frac{\overarc{AD} - \overarc{BC}}{2}\right) = 180 - \angle{BEC}. $ Therefore it suffices to show that $ \angle{XEY} = \angle{BEC}. $ But because $ \angle{FEC} = \angle{YEC}, $ it suffices to prove that $ EX $ and $ EF $ are isogonal conjugates with respect to $ \triangle{AED}. $ This is extremely easy with some trigonometry. Note that by Trig Ceva on $ F $ in $ \triangle{ADE} $ we have that $ \frac{\sin{AEF}}{\sin{DEF}} = \frac{\sin{ADF}}{\sin{EDF}} \cdot \frac{\sin{EAF}}{\sin{DAF}} = \frac{\sin{ADF}}{\sin{DAF}}. $ Similarly by Trig Ceva on $ X $ in triangle $ \triangle{BCE} $ we find that $ \frac{\sin{AEX}}{\sin{DEX}} = \frac{\sin{BCX}}{\sin{ECX}} \cdot \frac{\sin{EBX}}{\sin{CBX}} = \frac{\sin{BCX}}{\sin{CBX}} = \frac{\sin{DAF}}{\sin{ADF}} $ which implies that $ EX $ and $ EF $ are isogonal as desired.

While this solution may seem unmotivated, some insight into this configuration (Newton-Gauss Line) easily solves this problem. Let $P',Q'$ be the midpoints of $FB,FC$ respectively. Lemma: $\triangle NP'M\sim \triangle EDF.$ Proof: Note that $\triangle MP'Q'\sim\triangle FBC\sim \triangle FAD$ and $\triangle NP'Q'\sim\triangle EBC\sim \triangle EDA\implies EAFD\sim NQ'MP'$ which implies the desired. Lemma: $NGP'M$ cyclic. Proof: From the previous lemma, $\angle NMP'=\angle EFD.$ Then $\angle NGP'+\angle NMP'=\pi-\angle EGN-\angle BGP'+\angle EFD=2\pi-\angle EBF-\angle BEF-\angle EFB=\pi,$ as desired. Now $P'=P,Q'=Q$ so the conclusion follows. As an extension, show that if $FG\cap EC=G',FH\cap AB=H',$ the midpoint of $G'H'$ lies on $NM.$

If we name the midpoint of EC and BE , X and Y by using NG=NH , MG=MH we can proof that NMHX and NMGY are concyclic (because we get GH is perpendicular to NM ) and then it's easy to see if P' and Q' are midpoint of FB , FC then XNQ'H and YNP'G are concyclic and we get P'=P and Q'=Q .