[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(250); 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 = 4.276118500927452, xmax = 5.263967833975302, ymin = 2.15556608210655, ymax = 3.0356074215121707; /* image dimensions */ pen zzffff = rgb(0.6,1.,1.); /* draw figures */ draw((4.889779689274796,2.787258881791792)--(4.339686646478104,2.4020424507029134), linewidth(1.)); draw((4.339686646478104,2.4020424507029134)--(5.219663779074135,2.6334495960919755), linewidth(1.)); draw((4.889779689274796,2.787258881791792)--(5.219663779074135,2.6334495960919755), linewidth(1.)); draw((4.889779689274796,2.787258881791792)--(4.990804239747398,2.5732665177721805), linewidth(1.)); draw(circle((4.650564028974253,2.5434838264747026), 0.3415412269812749), linewidth(1.)); draw((4.339686646478104,2.4020424507029134)--(4.928494544786919,2.3449768036569134), linewidth(1.)); draw((4.824254395381144,2.7413731535676473)--(4.928494544786919,2.3449768036569134), linewidth(1.)); draw((4.928494544786919,2.3449768036569134)--(5.219663779074135,2.6334495960919755), linewidth(1.)); draw((5.219663779074135,2.6334495960919755)--(4.824254395381144,2.7413731535676473), linewidth(1.)); draw((4.990804239747398,2.5732665177721805)--(4.928494544786919,2.3449768036569134), linewidth(1.)); draw(circle((4.990804239747398,2.5732665177721805), 0.23664042692847906), linewidth(1.)); draw((4.889779689274796,2.787258881791792)--(4.928494544786919,2.3449768036569134), linewidth(1.)); /* dots and labels */ dot((4.339686646478104,2.4020424507029134),dotstyle); label("$A$", (4.344084409962773,2.413367806033629), NE * labelscalefactor); dot((4.889779689274796,2.787258881791792),dotstyle); label("$B$", (4.894842638352441,2.7988985659063976), NE * labelscalefactor); dot((5.219663779074135,2.6334495960919755),dotstyle); label("$C$", (5.22412574936839,2.6453893575680008), NE * labelscalefactor); dot((4.990804239747398,2.5732665177721805),dotstyle); label("$F$", (4.995619675887571,2.5844544046397817), NE * labelscalefactor); dot((4.928494544786919,2.3449768036569134),dotstyle); label("$D$", (4.933512896941503,2.3571201571768117), NE * labelscalefactor); dot((4.824254395381144,2.7413731535676473),dotstyle); label("$E$", (4.829220381352821,2.753197351210234), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); shipout(bbox(zzffff,Fill)); /* end of picture */[/asy][/asy] $(ABD)\cap AC=\{A,F\}$. We have $\angle AFB=\angle ADB=2\angle ACB\Rightarrow FB=FC$. $We have \angle BAD+\angle BDA+\angle ABD=180\Rightarrow BAD+2\angle BCA=120=2(\angle BAC+\angle BCA)\Rightarrow BAD=2\angle BAC\Rightarrow FB=FD$ We have $\angle AED=\frac{180-\angle BAD}2=\frac{BFD}2=\angle BCD\Rightarrow EBCD$ is concyclic$\Rightarrow FE=FB-FC=FD\Rightarrow \triangle EFC=\triangle DFC\Rightarrow EC=DC$.

Let $O$ be the circumcenter of $(ABC)$. We know that $\angle AOC=120$. $\angle ADE==2\angle C=\angle AOE \implies A,O,D,E$ are cyclic. $\implies \angle AOD=120$ So $O,D,C$ are collinear. $\angle EAC=60-\angle C$ Also $=60+\angle EDB=\angle DBE+\angle EDB=2\angle C-\angle EDB$ $\angle EDB=\angle C-30,\angle DEA= \angle C+30=\angle ADE\implies AC \perp ED$ $\implies CE=CD$ $30=\angle OCA= \angle DCA= \angle ACE \implies ECD$ is equilateral triangle so $EC=ED$