Let $\ell$ be the line through $F$ perpendicular to $KN$, and let $M$ (the midpoint of $AC$) be a variable point on $\ell$. As $M$ moves along $\ell$, $A$ moves along the line $\ell_A$ through $F$ with $\measuredangle(\ell, \ell_A) = 45^\circ$, and $C$ moves along the line $\ell_C$ through $F$ with $\measuredangle(\ell, \ell_C) = -45^\circ$. Let $B_1,B_2$ be the reflections of $A,C$ in $K,N$ respectively. As $A$ and $C$ move along their respective lines, $B_1$ and $B_2$ move along the reflections $\ell_A'$ and $\ell_C'$ of $\ell_A$ and $\ell_C$ in $K$ and $N$, respectively. The location of $B$ is the intersection of $\ell_A'$ and $\ell_C'$.

Let $P$ be the midpoint of $AC$. Note that $FP = \frac{1}{2}AC$, and by $NK$ midsegment, $FP = NK$. Hence, we can construct a perpendicular to $NK$ and mark a point $P$ beyond $F$ such that $FP = NK$. The triangle can be constructed from here. [asy][asy] import graph; size(8cm); 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 = -5.76, xmax = 17.2, ymin = -5.24, ymax = 10.04; /* image dimensions */ draw((1.7308925809959894,1.2658037292857292)--(1.5212945211769622,1.4557212201076646)--(1.331377030355027,1.2461231602886373)--(1.540975090174054,1.0562056694667021)--cycle, linewidth(1) + red); /* draw figures */ draw((0.08,2.38)--(2.21,0.45), linewidth(1)); draw(circle((5.64,5.58), 2.87433), linewidth(1)); draw((5.64,5.58)--(3.7100031596423135,3.450003487066387), linewidth(1)); draw((3.7100031596423135,3.450003487066387)--(1.540975090174054,1.0562056694667021), linewidth(1) + linetype("2 2") + red); /* dots and labels */ dot((5.64,5.58),linewidth(4pt) + dotstyle); label("$F$", (5.72,5.74), NE * labelscalefactor); dot((2.21,0.45),linewidth(4pt) + dotstyle); label("$K$", (2.28,0.6), NE * labelscalefactor); dot((0.08,2.38),linewidth(4pt) + dotstyle); label("$N$", (0.16,2.54), NE * labelscalefactor); dot((3.7100031596423135,3.450003487066387),linewidth(4pt) + dotstyle); label("$P$", (3.86,3.44), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Draw in $\triangle PNK$ and create parallelograms, taking each side to be the diagonal. That will restore $\triangle ABC$ [asy][asy] import graph; size(6cm); 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 = -5.76, xmax = 17.2, ymin = -5.24, ymax = 10.04; /* image dimensions */ pen ffwwqq = rgb(1,0.4,0); pen ffttww = rgb(1,0.2,0.4); pen ffccww = rgb(1,0.8,0.4); /* draw figures */ draw((0.08,2.38)--(2.21,0.45), linewidth(1) + ffwwqq); draw((0.08,2.38)--(3.7100031596423135,3.450003487066387), linewidth(1) + ffwwqq); draw((3.7100031596423135,3.450003487066387)--(2.21,0.45), linewidth(1) + ffwwqq); draw((2.21,0.45)--(5.840003159642313,1.5200034870663868), linewidth(1) + ffttww); draw((5.840003159642313,1.5200034870663868)--(3.7100031596423135,3.450003487066387), linewidth(1) + ffttww); draw((2.21,0.45)--(-1.4200031596423135,-0.6200034870663871), linewidth(1) + ffttww); draw((-1.4200031596423135,-0.6200034870663871)--(1.5800031596423134,5.380003487066387), linewidth(1) + ffttww); draw((1.5800031596423134,5.380003487066387)--(3.7100031596423135,3.450003487066387), linewidth(1) + ffttww); draw(circle((2.9600015798211565,1.9500017435331936), 2.911925067775071), linewidth(1)+ ffccww); draw(circle((1.145,1.415), 3.2742144708841856), linewidth(1)+ ffccww); draw(circle((1.8950015798211568,2.9150017435331934), 2.4850468004319053), linewidth(1) + ffccww); /* dots and labels */ dot((5.64,5.58),linewidth(4pt) + dotstyle); label("$F$", (5.72,5.74), NE * labelscalefactor); dot((2.21,0.45),linewidth(4pt) + dotstyle); label("$K$", (2.28,0.04), NE * labelscalefactor); dot((0.08,2.38),linewidth(4pt) + dotstyle); label("$N$", (0.16,2.54), NE * labelscalefactor); dot((3.7100031596423135,3.450003487066387),linewidth(4pt) + dotstyle); label("$P$", (3.86,3.44), NE * labelscalefactor); dot((5.840003159642313,1.5200034870663868),linewidth(4pt) + dotstyle); label("$A$", (6.08,1.6), NE * labelscalefactor); dot((-1.4200031596423135,-0.6200034870663871),linewidth(4pt) + dotstyle); label("$B$", (-1.9,-1.18), NE * labelscalefactor); dot((1.5800031596423134,5.380003487066387),linewidth(4pt) + dotstyle); label("$C$", (1.66,5.54), NE * labelscalefactor); dot((2.9600015798211565,1.9500017435331936),linewidth(4pt) + dotstyle); dot((1.145,1.415),linewidth(4pt) + dotstyle); dot((1.8950015798211568,2.9150017435331934),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Nice Problem! My first construction problem First draw $\ell$, the perpendicular from $F$ to $NK$ and then draw $\ell'$ a perpendicular to $\ell$ at $F$. Then draw the angle bisectors at $F$ of $\angle(\ell,\ell')$ Now construct a parallelogram $NKFX$ where $X$ lies on $\ell'$. Draw a circle centred at $F$ and with radius $FX$ and let it intersect $\ell$ at $P$. Now draw the circle centred at $P$ with radius $PF$ and let it intersect the angle bisectors at $A',C'$. These two points are the required $A$ and $C$ and $B$ can be obtained simply by reflecting $A$ over $N$ $\blacksquare$