Deux solutions

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.114cm); 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 = -12.82, xmax = 17.56, ymin = -6.25, ymax = 8.15; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); draw((-1.42,6.35)--(-7.54,-3.35)--(2.14,-3.33)--cycle, linewidth(0.7) + rvwvcq); draw((-7.115736836841633,-3.349123423216615)--(-7.116613413625019,-2.9248602600582485)--(-7.540876576783385,-2.925736836841634)--(-7.54,-3.35)--cycle, linewidth(0.7)); draw((2.139123423216615,-2.905736836841634)--(1.7148602600582488,-2.906613413625019)--(1.715736836841634,-3.330876576783385)--(2.14,-3.33)--cycle, linewidth(0.7)); draw((-1.400861465175194,-2.91305085520608)--(-1.8251246283335603,-2.913927431989465)--(-1.824248051550175,-3.338190595147831)--(-1.399984888391809,-3.337314018364446)--cycle, linewidth(0.7)); /* draw figures */ draw((-1.42,6.35)--(-7.54,-3.35), linewidth(0.7) + rvwvcq); draw((-7.54,-3.35)--(2.14,-3.33), linewidth(0.7) + rvwvcq); draw((2.14,-3.33)--(-1.42,6.35), linewidth(0.7) + rvwvcq); draw(circle((-2.7072050445591582,1.506817568342667), 5.0113184785097005), linewidth(0.7) + dtsfsf); draw((-7.552676403593336,2.785379339174353)--(-7.54,-3.35), linewidth(0.7) + dbwrru); draw((2.132691503792112,0.2073121646177605)--(2.14,-3.33), linewidth(0.7) + dbwrru); draw((-7.54,-3.35)--(2.132691503792112,0.2073121646177605), linewidth(0.7) + wvvxds); draw((-7.552676403593336,2.785379339174353)--(2.14,-3.33), linewidth(0.7) + wvvxds); draw((-1.42,6.35)--(-1.399984888391809,-3.337314018364446), linewidth(0.7) + wvvxds); /* dots and labels */ dot((-1.42,6.35),dotstyle); label("$A$", (-1.34,6.55), NE * labelscalefactor); dot((-7.54,-3.35),dotstyle); label("$B$", (-8.04,-3.29), NE * labelscalefactor); dot((2.14,-3.33),dotstyle); label("$C$", (2.22,-3.13), NE * labelscalefactor); dot((-1.399984888391809,-3.337314018364446),linewidth(4.pt) + dotstyle); label("$D$", (-1.18,-3.97), NE * labelscalefactor); dot((2.132691503792112,0.2073121646177605),linewidth(4.pt) + dotstyle); label("$N$", (2.46,0.53), NE * labelscalefactor); dot((-7.552676403593336,2.785379339174353),linewidth(4.pt) + dotstyle); label("$M$", (-8.12,2.61), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Toss the figure on the coordinate plane. Let $D = (0,0), \ A = (0,a), \ B = (b,0), \ C = (c,0)$. Since $\displaystyle \triangle ABC$ is acute, we may assume $a>0, b<0, c>0$. Now, $M$ has $x-$coordinate equal to $b$, and is on the line $y = \frac ab (x-c)$, thus $M = \left(b, \frac{b(b-c)}{a} \right)$. Symmetry in $b$ and $c$ implies that $N = \left(c, \frac{c(c-b)}{a} \right)$. The circumcentre of a triangle with vertices $(0,0), \ (0,a), \ (p,q)$ has coordinates $\left(\frac{p^2 + q^2 - aq}{2p}, \frac a2 \right)$. Now, $A,M,D,N$ are concyclic iff $\displaystyle \triangle AMD$ and $\displaystyle \triangle AND$ have the same circumcentre. Therefore, the problem asks us to show that $a = c-b$ if and only if $\left(\frac{a^2b + b(b-c)^2 - a^2(b-c)}{2a^2}, \frac a2 \right) = \left(\frac{a^2c + c(c-b)^2 - a^2(c-b)}{2a^2}, \frac a2 \right)$. But this is clear, since the difference of the $x-$coordinates is $\frac{1}{a^2}(c-b)((c-b)^2 - a^2)$ so this is $0$ iff $c-b = a$. (recall that $a>0, b<0, c>0$)

Observes $\triangle ADB \cong \triangle CBM$.Therefore we get $\angle BMD = \angle MDB = 45$ which give us $\angle MDA = 45$ Similarly we get $\angle NDA = 45$ which give us $\angle MDN = 90$. Now drop perpendicular from $M$ to $AD$ at $P$. we will get $MP = BD$ and $AP = DC$, similary $NQ \perp AD$ which give us $AQ= BD, NQ = DC$ Now as $\triangle APM \cong \triangle NQA$ we get $\angle MAN = 90$ hence $M,A,N,D$ cyclic. Second part we get $\triangle ADB \sim \triangle CBM$ and $\triangle ADC \sim \triangle BCN$ but as $\frac{AD}{BC}$ is always constant. We get $\triangle BMD \sim \triangle CND$ so $\angle BMD = \angle CND = \angle MDA = \angle NDA$ which give us $AD$ is angle bisector of $\angle MDN$ Now from $AM = AN$ we get $AD=BC$.