Let $Y,Z$ be the midpoints of $AC,AB.$ $G \equiv BY \cap CZ$ is the centroid of $\triangle ABC.$ If $CE$ cuts $(O_2)$ again at $L,$ then $\angle ELB=\angle EBA=\angle ECA$ $\Longrightarrow$ $AC \parallel BL$ $\Longrightarrow$ $GC:GL=GY:GB=-1:2$ $\Longrightarrow$ $\tfrac{GC \cdot GE}{GE \cdot GL}=-\tfrac{1}{2},$ i.e. the ratio of the powers of $G$ WRT $(O) \equiv \omega$ and $(O_2)$ equals $-\tfrac{1}{2},$ thus the center $U$ of $\odot(GBE)$ coaxal with $(O),(O_2)$ verifies $UO:UO_2=-1:2=GO:GH$ (H is the orthocenter of ABC) $\Longrightarrow$ $HO_2 \parallel GU,$ but $\angle UGE=90^{\circ}-\angle EBG=90^{\circ}-\angle DCG$ $\Longrightarrow$ $GU \perp CD$ $\Longrightarrow$ $HO_2 \perp CD$ $\Longrightarrow$ $HO_2 \parallel OO_1.$ Similarly $HO_1 \parallel OO_2$ $\Longrightarrow$ $OO_1HO_2$ is a parallelogram $\Longrightarrow$ $\overline{O_1O_2}$ and $\overline{OH}$ bisect each other through the 9-point center.

Let $ H $ be the orthocenter of $ \triangle{ABC} $. I shall show that $ O_{1}HO_{2}O $ is a parallelogram, which will immediately imply the desired result. We proceed with a complex number bash. Assume WLOG that the circumcircle of $ \triangle{ABC} $ is the unit circle and denote the complex coordinates of $ A, B, C $ by $ a, b, c $ respectively. Denote the complex coordinate of $ O_2 $ by $ x $. Then since $ O_{2}B \perp AB $ we have that $ \frac{x - b}{\overline{x} - \overline{b}} = -\frac{b - a}{\overline{b} - \overline{a}} = ab $ so $ x $ satisfies the equation $ x = ab\overline{x} + b - a $. Now, denote the complex coordinates of $ E $ by $ e $. Since $ E $ is on the $ C $-median of $ \triangle{ABC} $ we have that $ \frac{e - c}{\overline{e} - \overline{c}} = \frac{a + b - 2c}{\overline{a} + \overline{b} - 2\overline{c}} = \frac{(a + b - 2c)abc}{bc + ac - 2ab} $ so $ e $ satisfies the equation $ e = \frac{(a + b - 2c)abc}{bc + ac - 2ab}\overline{e} + \frac{(a + b)(c^2 - ab)}{bc + ac - 2ab} $ But $ e $ lies on the unit circle so $ \overline{e} = \frac{1}{e} $ and so, after plugging in, we find that $ e $ satisfies: $ e^2 - \frac{(a + b)(c^2 - ab)}{bc + ac - 2ab}e - \frac{(a + b - 2c)abc}{bc + ac - 2ab} = 0 $. Since $ c $ also is a root of this quadratic, by Vieta's formulas we find that $ e = \frac{(2c - a - b)ab}{bc + ac - 2ab} $. Now let $ M $ denote the midpoint of segment $ BE $ and let $ m $ be its complex coordinate. Then we easily find that $ m = \frac{b + e}{2} = \frac{b(3ac + bc - 3ab - a^2)}{2(bc + ac - 2ab)} $. Since $ O_{2}M \perp BE $ we have that $ \frac{x - m}{\overline{x} - \overline{m}} = -\frac{b - e}{\overline{b} - \overline{e}} $. But since $ E \in \omega $ we have that $ \overline{e} = \frac{1}{e} $ so $ \frac{x - m}{\overline{x} - \overline{m}} = be = \frac{(2c - a - b)ab^2}{bc + ac - 2ab} $. Therefore $ x = \frac{(2c - a - b)ab^2}{bc + ac - 2ab}\overline{x} + m - \overline{m} \cdot \frac{(2c - a - b)ab^2}{bc + ac - 2ab} $. But we can compute that $ \overline{m} = \frac{3ab + a^2 - 3ac - bc}{2ab(a + b - 2c)} $ so $ \overline{m} \cdot \frac{(2c - a - b)ab^2}{bc + ac - 2ab} = -\frac{b(3ab + a^2 - 3ac - bc)}{2(bc + ac - 2ab)} = m $ and so $ x = \frac{(2c - a - b)ab^2}{bc + ac - 2ab}\overline{x} $. Plugging this into our original equation $ x = ab\overline{x} + b - a $, we find that $ \frac{(2c - a - b)ab^2}{bc + ac - 2ab}\overline{x} = ab\overline{x} + b - a $ and solving for $ \overline{x} $ we obtain $ \overline{x} = \frac{ac + bc - 2ab}{ab(c - b)} $. Letting $ O_1 $ have complex coordinate $ y $ we similarly find that $ \overline{y} = \frac{ab + cb - 2ac}{ac(b - c)} $. Now since $ O $ has complex coordinate $ 0 $ and $ H $ has complex coordinate $ a + b + c $ it suffices to show that $ \overline{x} + \overline{y} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} $. But $ \overline{x} + \overline{y} = \frac{ac + bc - 2ab}{ab(c - b)} + \frac{ab + cb - 2ac}{ac(b - c)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} $ as desired so we are done.

that the midpoint of $\overline{BO'}$ is the nine-point center of $\triangle ABC.$ Meanwhile, since $O'O_1$ is the perpendicular bisector of $\overline{CU}$, we have \[\measuredangle UO_1O' = \measuredangle UDC = \measuredangle BDC = \measuredangle BAC \quad \text{and} \quad \measuredangle UO'O_1 = \measuredangle UAC = \measuredangle BCA,\]where the last step follows since $ABCU$ is a parallelogram. Therefore, $\triangle ABC \sim \triangle O_1UO'.$ Then since $O'U = R$ (which follows from the aforementioned homothety), we obtain $\tfrac{r_1}{AB} =\tfrac{O_1O'}{AC} = \tfrac{R}{BC}.$ Analagously, we find that $\tfrac{r_2}{AC} = \tfrac{R}{BC}$, and hence $O_1O' = r_2.$ Consequently, $O_1O' = O_2B$ and meanwhile $O_1O' \parallel O_2B$, because both lines are perpendicular to $AB.$ Therefore, $O_1O'O_2B$ is a parallelogram, and thus the common midpoint of $\overline{O_1O_2}$ and $\overline{BO'}$ is the nine-point center of $\triangle ABC.$ $\square$

Ok. I should attach a diagram

complex with a bit less computation

We dump on the complex plane. Let $a,b,c,d,e,z_1,z_2$ represent points $A,B,C,D,E,O_1,O_2$ and let $(ABC)$ be the unit circle centered at the origin. Then, we must have the midpoint of $AC$, or $\frac{a+c}{2}$, $D$ and $B$ be collinear. In other words: $$ \begin{vmatrix} \frac{a+c}{2} && \frac{\bar a + \bar c}{2} && 1 \\ b && \bar b && 1 \\ d && \bar d && 1 \end{vmatrix} =0$$Noting that $|d| =1$: $$ \begin{vmatrix} \frac{a+c}{2} && \frac{\bar a + \bar c}{2} && 1 \\ b && \bar b && 1 \\ d^2 && 1 && d \end{vmatrix} =0$$In other words $$d^2 \left( \frac{\bar a + \bar c - 2 \bar b}{2} \right) + \left( \frac{\bar b a + \bar b c}{2} - \frac{b \bar a + b \bar c}{2} \right)d - \frac{a+c-2b}{2} = 0$$Thus, the sum of the roots are: $$-\frac{\frac{\bar b a + \bar b c - b \bar a - b \bar c}{2}}{\frac{\bar a + \bar c - 2 \bar b}{2}} = -\frac{\frac{a+c}{2b} - \frac{bc + ab}{2ac}}{\frac{bc + ab + ab^2}{2abc}} = \frac{-a^2c-ac^2+b^2c+ab^2}{bc+ab-2ac}$$which after subtracting $b$ (the extraneous one) should yield $d$: $$\frac{-a^2c - ac^2 + 2abc}{bc + ab -2ac} = \frac{ac(-a-c+2b)}{bc + ab -2ac}$$We know that the segment from midpoint of $DC$ to $DC$ is perpendicular to $DC$, so: $$\frac{(\bar c - \bar d)z_1+(c-d)\overline{z_1} + \overline c d - c \overline d}{\overline c - \overline d} = c + d \Longleftrightarrow (\bar c - \bar d)z_1 = (d-c) \overline{z_1} \Longleftrightarrow z_1 = dc\overline{z_1}$$Note that since $CO_1$ is perpendicular to $AC$: $$\frac{z_1-c}{a-c} = \frac{\bar c - \overline{z_1}}{\bar a - \bar c} \Longleftrightarrow$$$$\frac{z_1-c}{a-c} = \frac{\bar c - \overline{z_1}}{\frac{1}{a} - \frac{1}{c}} \Longleftrightarrow$$$$\frac{z_1-c}{a-c} = \frac{da - dac \overline{z_1}}{d(c-a)} \Longleftrightarrow$$$$z_1-c = \frac{-da + z_1a}{d} \Longleftrightarrow$$$$z_1(d-a) = dc - da$$which means that: $$z_1 = \frac{c-a}{1 - \frac{a}{d}}$$Note that: $$\frac{a}{d} = \frac{a}{\frac{ac(-a-c+2b)}{bc + ab -2ac}} = \frac{1}{\frac{-ca-c^2+2bc}{bc + ab - 2ac}} = \frac{bc + ab - 2ac}{-ca -c^2 + 2bc}$$so $$1 - \frac{a}{d} = \frac{-ca -c^2 + 2bc - bc - ab + 2ac}{-ca -c^2 + 2bc} = \frac{ca - c^2 + bc - ab}{-ca - c^ 2 + 2bc} = \frac{(b-c)(c-a)}{-ca-c^2 + 2bc}$$from which we have that: $$z_1 = \frac{c-a}{\frac{(b-c)(c-a)}{-ca-c^2 + 2bc}} = \frac{-ca -c^2 + 2bc}{b-c}$$and by symmetry: $$z_2 = \frac{2bc - ab - b^2}{b-c}$$The midpoint of $O_1O_2$ is hence: $$\frac{-ca-c^2+ab+b^2}{c-b} = \frac{(a+b+c)(c-b)}{2(c-b)} = \frac{a+b+c}{2}$$which is precisely the nine point center, as desired.

Let $(ABC)$ be the unit circle in the complex plane and $M, N$ be the midpoints of $AC, AB$, respectively. We therefore denote the complex coordinates of $A, B, C, D, E, O_1, O_2, M, N$, as $a, b, c, d, e, o_1, o_2, m, n$, respectively. Using spiral similarities $D : CO_1 \rightarrow AO$ and $E : BO_2 \rightarrow AO$ ($AC$ is tangent to $(O_1)$ and $AB$ is tangent to $(O_2)$), we derive the following: (1) $o_1=\frac{c(a+c-2b)}{c-b}$ and (2) $o_2=\frac{b(a+b-2c)}{b-c}$ (we solved for $d$, $e$ using the equation $\frac{bd(a+c)-ac(b+d)}{bd-ac}=m$. Since $m$ is the midpoint, we have $\frac{bd(a+c)-ac(b+d)}{bd-ac}=m=\frac{a+c}{2}$) Adding (1) and (2), we have: $o_1+o_2=\frac{c(a+c-2b)}{c-b}+\frac{b(a+b-2c)}{b-c}$ $\Longleftrightarrow$ $o_1+o_2=\frac{b(a+b-2c)-c(a+c-2b)}{b-c}$ $\Longleftrightarrow$ $o_1+o_2=\frac{(b-c)(a+b+c)}{b-c}$, which gives $o_1+o_2=a+b+c=2n$, as desired $\blacksquare$

A quick fact I observed is that $N_9$ is the midpoint of $O_1O_2$.

Note that \begin{align*}c&=\frac{1}{2}(a+c+o_1-ac\overline{o_1}),\\ \frac{c+d}{2}&=\frac{1}{2}(c+d+o_1-cd\overline{o_1}),\\ \frac{a+c}{2}&=\frac{ac(b+d)-bd(a+c)}{ac-bd}.\end{align*}From the first equation, we have $o_1=c-a+ac\overline{o_1},$ from the second $\overline{o_1}=\frac{o_1}{cd},$ and from the third $d=\frac{ac(a-2b+c)}{-ab+2ac-bc}.$ Substituting from the second to the first, we have $o_1=c-a+ac\cdot\frac{o_1}{cd},$ or \begin{align*}o_1&=\frac{c-a}{1-\frac{a}{d}}\\&=\frac{c-a}{1-\frac{a}{\frac{ac(a-2b+c)}{-ab+2ac-bc}}}\\&=\frac{c-a}{1-\frac{-ab+2ac-bc}{c(a-2b+c)}}\\&=\frac{c-a}{\frac{ac-2bc+c^2+ab-2ac+bc}{ac-2bc+c^2}}\\&=\frac{c-a}{\frac{ab-bc-ac+c^2}{ac-2bc+c^2}}\\&=\frac{(c-a)(ac-2bc+c^2)}{(c-a)(c-b)}\\&=\frac{ac-2bc+c^2}{c-b}\\&=\frac{ac-2bc+c^2}{c-b}.\end{align*}Similarly, $$o_2=\frac{ab-2bc+b^2}{b-c}.$$Thus, \begin{align*}\frac{o_1+o_2}{2}&=\frac{ac-2bc+c^2-ab+2bc-b^2}{2(c-b)}\\&=\frac{ac+c^2-ab-b^2}{2(c-b)}\\&=\frac{(c-b)(c+b)+a(c-b)}{2(c-b)}\\&=\frac{(a+b+c)(c-b)}{2(c-b)}\\&=\frac{a+b+c}{2}\\&=n_9,\end{align*}so $O_1,O_2,N_9$ are collinear.

Use complex numbers with $(ABC)$ as the unit circle. Note that $O_1$ is the intersection of the perpendicular to $AC$ at $C$ and the perpendicular bisector of $BC$. The perpendicular to $AC$ at $C$ can be described as the line through $c$ and $-a$. Now, we will use an "abuse of notation" where we let $\sqrt{cd}$ denote an arbitrary complex number $x$ for which $x^2=cd$. Now, the perpendicular bisector of $CD$ intersects the unit circle at $\sqrt{cd}$ and $-\sqrt{cd}$ in some order. Thus, $O_1$ is the intersection of the chord through $-a$ and $c$ and the chord through $-\sqrt{cd}$ and $\sqrt{cd}$. By the unit circle intersection formula, this is $$\frac{d(c-a)}{d-a}.$$ Now of course we actually compute $d$. It is the second intersection of the line through $b$ and $\frac{a+c}{2}$ with the unit circle, so it is $$d=\frac{b-\frac{a+c}{2}}{b(\frac{\frac{1}{a}+\frac{1}{c}}{2})-1}=\frac{ac(2b-a-c)}{ab+bc-2ac}.$$Thus, $$o_1=\frac{d(c-a)}{d-a}=\frac{\frac{ac(2b-a-c)}{ab+bc-2ac}(c-a)}{\frac{ac(2b-a-c)}{ab+bc-2ac}-a}$$$$=\frac{c(2b-a-c)(c-a)}{c(2b-a-c)-(ab+bc-2ac)}=\frac{c(2b-a-c)(c-a)}{ac+bc-c^2-ab}$$$$\frac{c(2b-a-c)(c-a)}{-(c-a)(c-b)}=\frac{c(2b-a-c)}{b-c}.$$Similarly, $$O_2=\frac{b(2c-a-b)}{c-b}.$$We wish to show that these are collinear with $\frac{a+b+c}{2}.$ Multiplying by $2(b-c)$ it suffices to show that $$2c(2b-a-c),-2b(2c-a-b),(a+b+c)(b-c)$$are collinear. Expanding this out gives $$4bc-2ac-2c^2,-4bc+2ab+2b^2,ab+b^2-ac-c^2.$$We subtract $-4bc+2ac+2b^2$ from each to get $$8bc-2ac-2c^2-2ab-2b^2,0,ab+b^2-ac-c^2+4bc-2ab-2b^2.$$The first one is exactly twice the last one (after simplifying of course), so they are collinear and thus we are done. In fact, this shows that $N_9$ is the midpoint of $O_1O_2.$

Let $O$ be the circumcenter of $(ABC)$; we assume that $(ABC)$ is the unit circle. Then, note that $\triangle O_1CD \sim \triangle OAD$, hence $\frac{o_1-c}{c-d} = \frac{o-a}{a-d}$. Using the second intersection formula we can furthermore compute $$d= \frac{b-m}{b\overline m - 1} = \frac{2b-a-c}{\frac ba+\frac bc-2} = \frac{ac(2b-a-c)}{ab+bc-2ac}$$where $m = \frac{a+c}2$. It follows that $$o_1 = c+\frac{(-a)(c-d)}{a-d} = \frac{c(2b-a-c)}{b-c}.$$Computing the symmetric expression for $o_2$, we have $$o_1+o_2 = \frac{c(2b-a-c)}{b-c} + \frac{b(2c-a-b)}{c-b} = a+b+c$$hence the midpoint of $\overline{O_1O_2}$ is precisely the nine-point center.

Let $ABCP,ACBQ$ parallelograms and $K,L$ midpoints of $AC,AB$ then $KD\cdot KP=-KD\cdot KB=-KA\cdot KC=KC^2$ so $P\in(O_1)$ similarly $Q\in(O_2)$. Now let perpendicular bisectors of $BQ,CP$ meet at $X$ and let $A$ antipode be $Z$, now $XO_1ZO_2$ parallelogram. Midpoint of $XZ$ is the same as NPC when projected onto $AC$, since $\tfrac14(Q+B+2C)=\tfrac14(A+2B+C)$, and similarly on $AB$, so midpoint of $O_1O_2$ is also the NPC as desired.