Let us make edges $AD$ and $BC$ longer so they meet at a new point, DOG. We see that $180=\angle \textbf{DOG}ND+\angle \textbf{DOG}NC=2(\angle D+\angle C)$, so $\angle D+\angle C=90$, which means $\angle$ DOG must be a corner angle of ninety degrees. We observe that the small triangle $\textbf{DOG}AB$ is similar to the big triangle $\textbf{DOG}DC$ because $AB$ and $CD$ are parallel lines. We also see that the desired result is visible in the smaller triangle. Because these are parallel lines with a line crossing through them, we have $\angle \textbf{DOG}AB=\angle D$, $\angle \textbf{DOG}BA=\angle C$, $\angle \textbf{DOG}MA=\angle \textbf{DOG}ND,$, and $\angle \textbf{DOG}MB=\angle \textbf{DOG}NC$. Since such a line intersects $AB$ at its middle part $M$, then by similarity, $N$ must also have it intersect at its middle part.

Solution. Let the angle bisectors of $\angle MND$ and $\angle MNC$ meet $AD$ and $BC$ at $X$ and $Y$, respectively. $\angle ADN= \angle YNC$ and $\angle BCN=\angle XND \implies AD\parallel YN$ and $BC\parallel XN$, combining this with the fact that $AB\parallel CD \implies ADNY, BCNX$ are parallelograms$\implies DN=AY, NC=XB$, so $CN=ND\iff XM=MY$, so we need to show that $XM=MY$. We know that $\angle XNY=\angle XNM+\angle MNY= \frac{\angle MND+\angle MNC}{2}=90$. $\angle XNY=\angle XND=\angle NXM\implies MX=MN$, similarly $MN=MY\implies MX=MY\implies N$ is the midpoint of $CD.\blacksquare$

Let $T=AD\cap BC$. Let $x=\angle ADN$, and therefore $\angle MNC=2x$, also let $y=\angle BCN$, so $\angle MND=2y$. Now, note that \[x+y=\angle ADN +\angle BCN = \frac12\left(\angle MNC +\angle MND\right)=90\]Thus, $\angle DTC=90$. Since $M$ is the midpoint of $AB$, it follows that $M$ is the circumcenter of the right triangle $\triangle TAB$. Thus, \[\angle TMA = 2\angle TBA = 2\angle TCD =2\angle BCN =\angle MND\]Thus, it clearly follows that $T,M,N$ are collinear. Then, the homothety that take $\triangle TAB\to \triangle TDC$ takes $M\to N$, so $N$ is the midpoint of $CD$ and we're done. $\blacksquare$.

Unless I have missed something, this should be correct [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -18.533392768086504, xmax = 45.97856165660547, ymin = -19.80590020959896, ymax = 11.363858150793542; /* image dimensions */ pen ttttff = rgb(0.2,0.2,1); pen qqffff = rgb(0,1,1); /* draw figures */ draw((xmin, 1.7667883152895996*xmin + 9.304756338270597)--(xmax, 1.7667883152895996*xmax + 9.304756338270597), linewidth(2)); /* line */ draw((xmin, -0.5659987624697909*xmin + 5.760522161867692)--(xmax, -0.5659987624697909*xmax + 5.760522161867692), linewidth(2)); /* line */ draw((xmin, -0.015995062974205058*xmin-4.678106235538223)--(xmax, -0.015995062974205058*xmax-4.678106235538223), linewidth(2)); /* line */ draw((xmin, -0.015995062974205058*xmin-13.309519380619621)--(xmax, -0.015995062974205058*xmax-13.309519380619621), linewidth(2)); /* line */ draw((xmin, -0.5659987624697911*xmin-7.262845948275804)--(xmax, -0.5659987624697911*xmax-7.262845948275804), linewidth(3.2)); /* line */ draw((xmin, 1.7667883152895991*xmin-32.90922188963595)--(xmax, 1.7667883152895991*xmax-32.90922188963595), linewidth(3.2)); /* line */ draw((-7.843276274780103,-4.552652537599027)--(-12.684814091610802,-13.106624980408222), linewidth(3.2) + ttttff); draw((-7.843276274780103,-4.552652537599027)--(18.979196698093652,-4.981679681924056), linewidth(3.2) + ttttff); draw((18.979196698093652,-4.981679681924056)--(34.67256958448942,-13.864109314601034), linewidth(3.2) + ttttff); draw((-12.684814091610802,-13.106624980408222)--(34.67256958448942,-13.864109314601034), linewidth(2.8) + ttttff); draw((-1.5193131898720451,6.620451547139299)--(-7.843276274780103,-4.552652537599027), linewidth(3.2) + qqffff); draw((-1.5193131898720451,6.620451547139299)--(18.979196698093652,-4.981679681924056), linewidth(3.2) + qqffff); draw((-1.5193131898720451,6.620451547139299)--(10.99387774643931,-13.485367147504629), linewidth(3.2) + red); /* dots and labels */ dot((-1.5193131898720451,6.620451547139299),dotstyle); label("$K$", (-0.8705296971971921,6.735621303341321), NE * labelscalefactor); dot((-7.843276274780103,-4.552652537599027),dotstyle); label("$A$", (-8.662969287295418,-3.984886496369432), NE * labelscalefactor); dot((18.979196698093652,-4.981679681924056),dotstyle); label("$B$", (19.059224890811603,-4.551609375649296), NE * labelscalefactor); dot((34.67256958448942,-13.864109314601034),dotstyle); label("$C$", (34.88023860404133,-13.383040911093838), NE * labelscalefactor); dot((5.567960211656775,-4.767166109761542),linewidth(4pt) + dotstyle); label("$M$", (5.7412372277346355,-4.40992865582933), NE * labelscalefactor); dot((-12.684814091610802,-13.106624980408222),linewidth(4pt) + dotstyle); label("$D$", (-12.535575629041203,-13.996990696980358), NE * labelscalefactor); dot((10.99387774643931,-13.485367147504629),linewidth(4pt) + dotstyle); label("$N$", (10.88897004786013,-14.563713576260222), NE * labelscalefactor); dot((16.57662819730869,-3.6218288837308665),linewidth(4pt) + dotstyle); label("$E$", (17.35905625297199,-3.2292559906629474), NE * labelscalefactor); dot((-7.102063640741422,-3.24308671663446),linewidth(4pt) + dotstyle); label("$F$", (-7.48229662212902,-2.3791716717431517), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\overleftrightarrow{AD} \cap \overleftrightarrow{BC} = {K}$ Let $\overline{NE}$ and $\overline{NF}$ be angle-bisectors of $\angle{MNC}$ and $\angle{MND}$ respectively Observe that $\angle{FNM} + \angle{ENM} = \frac{\angle{CNM} + \angle{DNM}}{2} = 90^{\circ}$ Since $\angle{CNE} = \angle{CDK}$, $KD \parallel EN$ Similarly $KC \parallel FN$ The above 2 statements imply $\angle{DKC} = 90^{\circ} = \angle{AKB}$ This along with the fact that $M$ is the midpoint of $AB$ imply that $M$ is the circumcircle of $\triangle KAB$ $\implies MA = MK = MB$ $\angle{NCK} = \angle{MBK} = \angle{MKB}$ $\implies NC = NK$ Similarly $ND = NK$ Together both imply - NC = ND Hence Proved!

Pretty simple for IGO (Unless my solution is wrong )

Here's my solution: By angle chase, we can conclude that $\angle ADC+\angle BCD=90^\circ$ Now, let $\overrightarrow{DA} \cap \overrightarrow{NM}=E$ $\rightarrow$ $\angle MAE=\angle ADC$ and $\angle EMA=\angle MND=2\angle BCD=180-2\angle ADC$. Hence $\angle MAE=\angle MEA$ and $ME=MA=MB$. $\angle ABC= 180^\circ -\angle BCD = 90^\circ +\angle ADC$. Also, $\angle MBE=\frac{180^\circ-\angle EMB}{2}= \frac{180^\circ-\angle MNC}{2}= \frac{180^\circ-2\angle ADC}{2}=90^\circ -\angle ADC$. So, $\angle ABC +\angle MBE = 180^\circ \rightarrow$ Lines $AD,MN,$ and $BC$ are concurrent. So, by similarity, we can have $DN=NC$.

Here is my solution Enough that $DA,NM,CB$ concurrent Let $L=DA \cap NM$ Then $DN=NL, \implies AM=ML$ but $AM=MB$ $\implies \angle DLC = 90º$ In the cuadrilateral $DLBC \implies \angle LBC=180º$ $L,B,C$ are collineals

Let $AD \cap BC = Z$. The existence of such a point $N$ implies that $\angle{DZC}=90$. Let $\angle{ADC}=x$. We get that $\angle{ABZ}=90-x$, and since $M$ is the midpoint of the hypotenuse in $\triangle{AZB}$, $\angle{ZMB}=2x, \angle{NMB}=180-2x$ (the latter by parallel lines). Therefore $Z,M,N$ are collinear, implying the desired

Construct a parallelograms $AMNA'$ and $BMNB'$. Observe that $A'N=AM=BM=B'N$. By the condition of the problem we have that $DA'=A'A=MN=BB'=B'C$, hence $DN=DA'+A'N=CB'+B'N=NC$, as needed.

Let $A'=AD\cap BC$. From the angle condition we have $\angle ADC+\angle BCD=90^\circ$, therefore $\angle DA'C=90^\circ$. Therefore $\triangle A'MB$ is isosceles. Notice \[ \angle BMA'+\angle BMN=\angle 180^\circ-2\angle BCD+2\angle BCD=180^\circ \]So $A',M,N$ are collinear. Since $M$ is the midpoint therefore $N$ is also the midpoint of $BC$.

Extend $AD$ and $BC$ to meet at $E$. Note that what we want to show is equivalent to $E,M,N$ being collinear. Let $\angle ADN=\alpha$, so $\angle MNC=2\alpha$. Then, $\angle MND=180-2\alpha$, so $\angle BCN=90-\alpha$. Therefore, $\angle DEC=90.$ Therefore, $M$ is the circumcenter of $\triangle AEB$, so $\angle EAM=\alpha$, $\angle AEM=\alpha$, so $\angle EMA=180-2\alpha$. Additionally, $\angle DAM=180-\alpha$, so from quadrilateral $AMND$, $\angle AMN=2\alpha$. Since $\angle EMA+\angle AMN=180$, $E,M,N$ are collinear, so we are done.

We can prove that $AD, BC, MN$ concur, by extending MN to meet AD and BC at $T$ and $T'$ and showing $MT = MT'$. From here the result follows by homothety.