Let $\triangle ABC$ and $\triangle A'B'C'$ be two triangles with the same circumcircle $(O,R).$ $(N,\frac{_R}{^2})$ is 9 point circle of $\triangle ABC$ and assume that midpoints $E',F'$ of $\overline{C'A'},\overline{A'B'}$ are on $(N).$ There are only two distinct circles with radius $\frac{_1}{^2}R$ passing through $E',F',$ namely, $\odot(A'E'F),$ because $\triangle A'E'F'$ is image of $\triangle A'B'C'$ under homothety $(A',\frac{_1}{^2}),$ and the 9-point circle of $\triangle A'B'C'$ $\Longrightarrow$ $(N)$ coincides with the 9-point circle of $\triangle A'B'C'.$ P.S. $\triangle ABC, \triangle A'B'C'$ is a poristic family with fixed circumcircle $(O)$ and MacBeath inconic (conic with focus O and pedal circle (N)).

suppose both the triangles have the unit circle centered at origin as their circumcircle. suppose $a,b,c$ be the vertices of the first triangle and that of the 2nd one are $d,e,f$ suppose the nine point circle of the first triangles contains the midpoints of 2 sides $DE ,DF$ then the equation of the nine point circle of the 1st triangle is $\mid {z-(a+b+c)/2}=1/2$ then $\mid{d+e-a-b-c}\mid=1$ and $\mid{d+f-a-b-c}\mid=1$. suppose $a+b+c=d+e+f+p$ . then $\mid{f+p}\mid=1$ and $\mid{e+p}\mid=1$. so , $p=0$ or ,$p=-e-f$ .note that , $p=-e-f$ implies $a+b+c=d$ , ,which implies that , the orthocentre of $\Delta ABC$ is $D$ ,which is impossible. so , $p=0$ ,so orthocentre of both the triangles is the same. so , the nine point centre and the radius of the nine point circle of both triangles is the same. hence done

mathbuzz wrote: then $\mid{f+p}\mid=1$ and $\mid{e+p}\mid=1$. so , $p=0$ or ,$p=-e-f$ . I'm really sorry, but I don't understand why.

Perform an inversion that leaves circle $(O)$ invariant. The hypothesis reduces to the following: there are given two triangles, which share the same incircle and 5 out of 6 vertices lie on the same circle. By Poncelet porism, the sixth vertice will also lie on the circle, which (by 'undoing' the inversion) proves that the third midpoint of the second triangle also lie on the $9$ point circle, meaning that the two triangle share the same Euler circle.

Let $\triangle{ABC}$ and $\triangle{XYZ}$ have the same circumcircle $\omega$ with radius $R$ and $H=orthocenter$ $of$ $\triangle{ABC}$ and ,$\gamma=$the Euler circle of $\triangle{ABC}$.Suppose $\gamma$ passes through midpoints of $XY$ and $XZ$. It is well known that for any point $W$ on $\omega$ the midpoint of $HW$ is on $\gamma$,hence applying this to point $X,Y,Z$ we find that the midpoints of $XH,YH,ZH$ are on $\gamma$.Thus $\gamma$ is the Euler circle of triangles:$\triangle{HXY}$ and $\triangle{HXZ}$ so $R_{XYH}=R_{XZH}=R$.Let $H'=orthocenter$ of $\triangle{XYZ}$.The only 2 circles passing through $X,Y$ with radius $R$ are $\omega$ and $(H'XY)$.H' is not on $\omega$($\triangle{XYZ}$ is acute angled) so $H'\in(HXY)$,in a similar manner we get $H'\in(HXZ)$,but H,X,H' are on both circles so H=H'.Two triangles with the same orthocenter and circumcenter and with the same circumradius have the same Euler circle.

Let $\triangle{ABC}$ have circumradius $R$, let $M_Y$, $M_Z$ be the midpoints of $XZ, XY$, and let the nine-point circle of $\triangle{ABC}$ be $\omega$. It is well known that the nine-point radius of any triangle is half its circumradius (this can be also seen by looking at the Euler line under $h(H, 1/2)$), and there are exactly two circles with radius $R/2$ that pass through $M_Y, M_Z$. But note that $(XM_YM_Z)$ is such a circle but it is clearly not the nine-point circle of $\triangle{XYZ}$, so thus $\omega$ is the nine-point circle of $\triangle{XYZ}$, as desired/

Suppose that $R$ is the circumradius of $\triangle ABC$. Now there are only two circles with radius $R/2$ passing through the midpoints of $\overline{XY}$ and $\overline{YZ}$; namely, the nine-point circle of $\triangle XYZ$ and the circle passing through $X$ and the midpoints of $\overline{XY}$ and $\overline{XZ}$. But this latter case clearly doesn't work, as it implies the nine-point circle of an acute triangle $\triangle ABC$ intersects the circumcircle of $\triangle ABC$, absurd.

Lets fix point $X$. Let the midpoints of $XY$ and $XZ$ be $M,N$. We know that both $M,N$ lie on the circle with diameter $OX$, where $O$ is the circumcenter of $(ABC)$, and on $\gamma $. Now let $H$ be the second point of intersection of $XY$ and $\gamma $. Claim: $NH=NX$ Proof: The circle with diameter $OX$ and $\gamma $ have equal diameters, and $M$ and $N$ are their intersection points, thus $\angle MHN = 180 - \angle MXN$ implying that $XN=HN$. From this (by a homothety with center $X$ and factor 2) we get that $H$ is the foot of altitude of $Z$ to $XY$, implying that $\gamma $ is the nine-point circle of triangle $XYZ$, which gives us the desired result.

The problem is equivalent to showing that the nine-point circles of $ABC$ and $XYZ$ coincide, or that the two triangles share an orthocenter. Thus, it suffices to show the following subproblem: Subproblem. Let $H$ be a point inside an acute triangle $ABC$ such that the reflections of $H$ over the midpoints of $\overline{AB}, \overline{AC}$ lie on the circumcircle. Show that $H$ is the orthocenter. But this is clearly evident because there exists a unique chord of the circumcenter parallel to the midline $\overline{MN}$ that is not $\overline{BC}$, and $H$ clearly works.

Let $\gamma'$ be the nine point circle of $XYZ$. We'll prove that $\gamma = \gamma'$, showing that $\gamma$ passes through the midpoint of $YZ$. Assume the contrary, $\gamma \neq \gamma'$. Let $M$ be the midpoint of $YZ$. Then $\gamma'$ passes through $M$. Since $\gamma$ and $\gamma'$ are congruent circles both pass through the midpoints of $XY$ and $XZ$, so $\gamma$ is the reflection of $\gamma'$ wrt line passing through the midpoints of $XY$ and $XZ$. Hence $\gamma$ passes through $X$, a contradiction. Therefore $\gamma = \gamma'$ implies $\gamma$ passes through the midpoint of $YZ$, as desired. $\blacksquare$

Notice both $9$ point circles pass through the midpoint of $\overline{XY}$ and $\overline{XZ}$. Since the circumcirlce is the same, they both have the same radius, so they must be the same circle, as desired. Let $N_A$ be the center of the nine-point circle of $\Delta ABC$ and $N_X$ of $\Delta XYZ$. Thus the two circles are the same, which finishes.

Weird problem didn't feel comfortable at all. I hope this solution works. Consider the homothety $\mathcal{H}_1$ with scale factor $k_1=2$ mapping the nine point circle of $\triangle ABC$ to its circumcircle $\Gamma$, and the homothety $\mathcal{H}_2$ with scale factor $k_2=\frac{1}{2}$ mapping $\Gamma$ to the nine point circle of $\triangle XYZ$. Since $k_1k_2=1$, we know that the composition of homotheties $\mathcal{H}_1$ and $\mathcal{H}_2$, which maps the nine point circle of $\triangle ABC$ to the nine point circle of $\triangle XYZ$ must be a translation. Let $M$ and $N$ denote the midpoints of $XY$ and $XZ$. Since the nine point circles of $\triangle ABC$ and $\triangle XYZ$ are translations of each other, they must be congurent, and thus, chord $MN$ subtends an equal angle in both these nine point circles. Since $YZ=2MN$ and the ratio between the radii of the nine-point circle of $\triangle ABC$ and $\Gamma$ is 2, it is not hard to see that the angle $(\theta)$ subtended by chord $EF$ on $\Gamma$ is equal to the arc subtended by $MN$ on the nine-point circle of $\triangle ABC$. Let $L$ be the midpoint of $YZ$. Then, $XMLN$ is a parallelogram, so \[\angle MLN = \angle MXN = \angle YXZ = \theta\]which is precisely the angle subtended by chord $MN$ on the nine point circle of $\triangle ABC$, which implies that $L$ must also lie on the nine point circle of $\triangle ABC$, as desired.

With respect to $\Delta XYZ$, the $NPC$ of $\Delta ABC$ is the circle which passes through two of its midpoints, has a radius equal to half of its circumradius and does not intersect its circumcircle. Now there are exactly two circles satisfying the first two properties, one of them being the nine point circle of $\Delta XYZ$ and the other passing through $X$, which cannot happen, so the nine point circle of $\Delta ABC$ must be the nine point circle of $\Delta XYZ$.