Let $D,E,F$ be the midpoints of $BC,CA,AB.$ Circle $(A')$ through $D$ and the projections of $B,C$ on internal bisectors of $\angle ACB,$ $\angle ABC$ is 9-point circle of $\triangle IBC,$ where $I$ is incenter of $\triangle ABC$ and similarly $(B'),(C')$ are 9-point circles of $\triangle ICA,$ $\triangle IAB.$ They concur then at the Poncelet point $Fe$ of $ABCI;$ Feuerbach point of $\triangle ABC.$ Let $X,Y,Z$ be the midpoints of $IA,IB,IC$ and let $N$ be the 9-point center of $\triangle ABC.$ We have $NB' \perp FeE$ and $B'C' \perp FeX$ $\Longrightarrow$ $\angle NB'C'=\angle XFeE$ and similarly $\angle NC'B'=\angle XFeF.$ But since $FeX$ bisects $\angle EFeF$ (well-known) then $\angle NB'C'=\angle NC'B'$ $\Longrightarrow$ $NB'=NC'$ and similarly we'll get $NC'=NA'$ $\Longrightarrow$ $N$ is circumcenter of $\triangle A'B'C'.$

For a more general configuration see Nine point center lies on the Euler line (lemma at post #3). Additionally, the incenter I of ABC is orthocenter of A'B'C'. For a proof See About 9-point centers.

ComplexPhi wrote: Let $ABC$ be a triangle . Let $A'$ be the center of the circle through the midpoint of the side $BC$ and the orthogonal projections of $B$ and $C$ on the lines of support of the internal bisectrices of the angles $ACB$ and $ABC$ , respectively ; the points $B'$ and $C'$ are defined similarly . Prove that the nine-point circle of the triangle $ABC$ and the circumcircle of $A'B'C'$ are concentric. My solution. Lemma. Given $\triangle{ABC}$ and $O, N$ are its circumcenter and nine-point center. $M = R_{BC}(O)$ then $N$ is the midpoint of $AM$ Proof. Too trivial Back to our main problem. Let $I$ be the incenter of $\triangle{ABC}$ $D, E, F$ are the midpoints of $\overarc{BC}, \overarc{CA}, \overarc{AB}$ $X, Y, Z$ are the reflections of $D, E, F$ WRT $BC, CA, AB$ It’s clear that $D$ is the circumcenter of $\triangle{IBC}$, $X = R_{BC}(D), A’$ is the nine-point center of $\triangle{IBC}$ $\Rightarrow A’$ is the midpoint of $IX$ (lemma) Analogously, we have: $H_I^2: A’\mapsto{X}, B’\mapsto{Y}, C’\mapsto{Z}$ (1) $\Rightarrow$ Center of $(A’B’C’) \mapsto$ center of $(XYZ)$ But where is the center of $(XYZ)$? We must remember that $(XYZ)$ is the Fuhrmann circle of $\triangle{ABC}$, and $HN_a$ is its diameter, i.e the center of $(XYZ)$ is the midpoint $S$ of $HN_a$! Notice that the centroid $G$ of $\triangle{ABC}$ lies on $IN_a$ s.t. $\frac{GI}{GN_a} = \frac{-1}{2}$. Moreover, the nine-point center $N$ of $\triangle{ABC}$ lies on $HG$ s.t. $\frac{NH}{NG} = -3$ (as you can see in the figure) Because of all that, $N$ must be the midpoint of $IS$, then due to the homothety (1), $N$ is the center of $(A’B’C’)$. Q.E.D

Use the right angles on the in-touch chord lemma(see Yufei Zhao) to observe that $A'$ is the nine point center of $\triangle BIC$. Now just complex bash it(pretty straight forward.) P.S.- The complex bash is similar to the solution given below.

Back to the problem at hand, note that $A'$ is just the nine-point center of $\triangle BIC$, where $I$ is the incenter. Now, let $M, N, P$ be the midpoints of arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ on $\odot (ABC).$ It is well-known that $M$ is the circumcenter of $\triangle BIC.$ Thus, if $M'$ is the reflection of $M$ in $BC$, it follows from the lemma that $A'$ is the midpoint of $\overline{IM'}.$ Now, we use complex numbers. WLOG set $\odot (ABC)$ to be the unit circle. It is well-known that there exist complex numbers $x, y, z$ such that \[a = x^2, b = y^2, c = z^2, m = -yz, n = -zx, p = -xy, i = -xy - yz - zx.\]Thus, we compute \[a' = \frac{i + m'}{2} = \frac{i + \left(b + c - bc\overline{m}\right)}{2} = \frac{y^2 + z^2 - x(y + z)}{2}.\]Then if $Q$ is the nine-point center of $\triangle ABC$, we have $q = \tfrac{x^2 + y^2 + z^2}{2}$, implying that \[\left|q - a'\right| = \left|\frac{x(x + y + z)}{2}\right| = \frac{1}{2}|x + y + z|,\]which is symmetric in $x, y, z.$ Therefore, $\left|q - a'\right| = \left|q - b'\right| = \left|q - c'\right|$, implying that $Q$ is the circumcenter of $\triangle A'B'C'$, as desired. $\square$

ComplexPhi wrote: Let $ABC$ be a triangle . Let $A'$ be the center of the circle through the midpoint of the side $BC$ and the orthogonal projections of $B$ and $C$ on the lines of support of the internal bisectrices of the angles $ACB$ and $ABC$ , respectively ; the points $B'$ and $C'$ are defined similarly . Prove that the nine-point circle of the triangle $ABC$ and the circumcircle of $A'B'C'$ are concentric. Just use homothety centred at G with ratio -1/2. Will add details soon.

Let $I, G, O, N$ be the incenter, centroid, circumcenter, nine-point center of $\triangle ABC.$ Let $G'$ be the centroid of $\triangle BIC$, and note that $A'$ is just the nine-point center of $\triangle BIC.$ Let $M$ and $D$ be the midpoints $\overline{BC}$ and arc $\widehat{BC}$ on $\odot(ABC).$ The homothety $\mathbf{H}(G, -2)$ sends $\overline{MN} \mapsto \overline{AO}.$ Similarly, since $D$ is the circumcenter of $\triangle BIC$ (well-known), the homothety $\mathbf{H}(G', -2)$ sends $\overline{MA'} \mapsto \overline{ID}.$ Thus, if $J$ is the reflection of $I$ in the midpoint of $\overline{AD}$, we have $AJ = ID = 2 \cdot MA'.$ Moreover, $AO = 2 \cdot MN$ and $\angle(AO, AJ) = \angle(MN, MA')$ since homothety preserves angles. This is enough to imply that $\triangle AOJ \sim \triangle MNA'$ with similarity coefficient $2.$ Therefore, $NA' = \tfrac{1}{2} \cdot OJ = \tfrac{1}{2} \cdot OI$, where the last step follows from symmetry in the perpendicular bisector of $\overline{AD}.$ Writing analagous relations, we obtain $NA' = NB' = NC'$, as desired. $\square$

$\odot A'B'C'$ is the image of Fuhrmann's circle under $\mathcal{H}_{I,\frac{1}{2}}$.However the center of the Fuhrman's circle is the midpoint of $\overline{HN_a}$ and now using $\overline{IN_a}=2\overline{IG}$ we obtain the desired.$\blacksquare$