Any solution?

Can be done easily with Cartesian coordinates

İs there a synthetic solution?

use law of sines in $BED$ and $BEC$ to prove $BECD$ is cyclic.

Or else just let $E' = \odot (BDC) \cap AB$. Now,$\angle E'BC=\angle E'DC=60^{\circ}$. Also, $\angle E'BD =120^{\circ}$ $\implies$ $\angle E'CD =60^{\circ}$ $\implies$ $\angle E'DC=\angle E'CD$. Hence, $E'$ lies on perpendicular bisector of $CD$. But this point must be $E$. Thus, $ECD$ is equilateral

Take $F\in BC|\triangle EBF$ is equilateral; triangles $BDE$ and $FCE$ are congruent according to 4th case of triangle congruency (obtuse-angled triangles), thus $\angle BDE=\angle BCE, BDCE$ is cyclic, done. Remark: if $E$ is outside segment $AB$, take then $F\in BD$. Best regards, sunken rock

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -6.28, xmax = 18.2, ymin = -5.76, ymax = 7.4; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((0.82,1.74)--(5,1.82), linewidth(2) + wrwrwr); draw((xmin, -1.6579521036448839*xmin + 3.0995207249888046)--(xmax, -1.6579521036448839*xmax + 3.0995207249888046), linewidth(2) + wrwrwr); /* line */ draw((0.82,1.74)--(2.840717967697245,5.3999861878189535), linewidth(2) + wrwrwr); draw((2.840717967697245,5.3999861878189535)--(5,1.82), linewidth(2) + wrwrwr); draw((5,1.82)--(2.979282032302755,-1.8399861878189534), linewidth(2) + wrwrwr); draw((5,1.82)--(1.4683634535646495,0.6650444482360276), linewidth(2) + wrwrwr); draw((2.2339608787128515,4.301009190264568)--(5,1.82), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.82,1.74),dotstyle); label("$B$", (0.9,1.94), NE * labelscalefactor); dot((5,1.82),dotstyle); label("$C$", (5.08,2.02), NE * labelscalefactor); dot((2.840717967697245,5.3999861878189535),linewidth(4pt) + dotstyle); label("$A$", (2.92,5.56), NE * labelscalefactor); dot((2.979282032302755,-1.8399861878189534),linewidth(4pt) + dotstyle); label("$A'$", (3.06,-1.68), NE * labelscalefactor); dot((1.4683634535646495,0.6650444482360276),dotstyle); label("$D$", (1.54,0.86), NE * labelscalefactor); dot((2.2339608787128515,4.301009190264568),linewidth(4pt) + dotstyle); label("$D'$", (2.32,4.46), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] let the symmetric point of $A$ against $BC$ : $A'$. $\triangle ABC \equiv \triangle BA'C$ , and let $D'$ be the corresponding point of $D$. $\Rightarrow$ $\triangle CDD'$ is an equilateral triangle , $CD' = DD'$ $\Rightarrow$ $D' = E$, and we are done

see the figure:

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