Some one help solve it pls...

$AX,BY,CZ$: altitudes, $\omega$: NPC, $N$: NPC center $J,K,L,Q,R,U,V$: midpoints of $AM,BE,CF,AC,AB,TS,AH$ resp. Denote the radius of the circumcircle by $R$ and $MB=MC=x$ Since $\overrightarrow{BY} = \overrightarrow{ME}$, $BMEY$ is a parallelogram. Hence, $K$ is the midpoint of $MY$. Similarly, $L$ is the midpoint of $MZ$. Also, $J$ is the midpoint of $QR$. Note that $QR,MY,MZ$ are chords of $\omega$ and $QR=MY=MZ=x/2$. Therefore, $NJ=NK=NL=\sqrt{R^2-x^2}/2$. ($\because$ The radius of $\omega$ is $R/2$.) Since $\angle HMX=\angle DMX=\angle UXM$ and $\angle NMX=\angle NXM$, we get $\angle NMH = \angle NXU=\theta$. By cosine's law on $\triangle MHV$, \[R^2 + MH^2 -2R\cdot MH \cdot \cos\theta = VH^2 = R^2 -x^2 \implies MH^2 - (2R\cos\theta)\cdot MH + x^2 = 0 \quad\cdots (1)\]By cosine's law on $\triangle XUN$, \[(R/2)^2+XU^2 - 2\cdot R/2 \cdot XU\cdot \cos\theta = NU^2 \implies (2\cdot NU)^2 = R^2 + MT^2 - 2R\cdot MT \cdot\cos\theta \quad\cdots (2)\]By (1) and $MT=x^2/MD=x^2/MH$, we get \[MT^2 - (2R\cos\theta)\cdot MT + x^2 = 0 \quad\cdots (3)\]Combining (2), (3) gives $(2\cdot NU)^2 = R^2 -x^2$. Therefore, $NU=NJ=NK=NL$!

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My solution: Let $AQ, BN, CP$ be the three altitudes of triangle $ABC$. Let $A'$ be the antipode of $A$, $A'M \cap \odot(ABC) = \lbrace A'; Z \rbrace$. Denote $I, J, K, G$ be the midpoints of segments $CF, ST, MN, MA$, respectively. We easily have $\overrightarrow{MP} = \overrightarrow{BF} = 2\overrightarrow{MI}$. Then, $I$ is the midpoint of segment $MP$. Similarly, $K$ is the midpoint of segment $MN$. Besides, since $\angle BAD = 90^o - \angle ABC = 90^o - \angle AA'C = \angle A'AC$. Therefore, $A'D \parallel BC$. Hence $ZT \parallel DA' \parallel ZT$. Let $Y$ is the reflection of $Z$ in $AH$. Since $BHCA'$ is the parallelogram, $H, M, A, Z$ are collinear. Therefore, $S, H, Y$ are collinear, and $A, Z, H, Y, P, N$ lie on the circle with diameter $AH$. Moreover, since $SHMD$ is a rhombus, $SH \parallel DM$. Hence, $SYTM$ is the parallelogram. Then, $\overrightarrow{MY} = 2\overrightarrow{MJ}$. Let $\Phi$ is the homothety with center $M$ and ratio $\dfrac{1}{2}$. We have $$ \Phi: P \mapsto I, N \mapsto K, A \mapsto G, Y \mapsto J$$Since $P, N, A, Y$ are concyclic, $I, K, G, J$ are concyclic.