$\definecolor{A}{RGB}{121,96,253}\color{A}\fbox{Solution}$ Let $M$ be the midpoint of $BC$ and $F\neq A$ be the intersection of line $AD$ with circumcircle of $ABC$. Observe that$$\angle EFC=\angle ABM,\ \angle FEC=\angle DEC=\angle BAM\implies \triangle BAM\sim\triangle FEC\implies \frac{CF}{EF}=\frac{BM}{AB}.$$Also $$\angle BEF=\angle BED=\angle BAC,\ \angle BFE=\angle BCA\implies \triangle BFE\sim\triangle BCA\implies \frac{EF}{BF}=\frac{AC}{BC}.$$Since $AB=AC$ half-line $FD^\rightarrow$ is angle bisector of $BFC$. Therefore $$\frac{CD}{BD}=\frac{CF}{BF}=\frac{CF}{EF}\cdot\frac{EF}{BF}=\frac{BM}{AB}\cdot \frac{AC}{BC}=\frac12.\blacksquare$$[asy][asy]import graph; size(5cm); 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 = -12.631253951432154, xmax = 2.3780768291613446, ymin = -2.2100169179008247, ymax = 6.6973061703442225; /* image dimensions */ pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen ffqqff = rgb(1,0,1); /* draw figures */ draw((-8.743526636724253,0.5744374818469364)--(-1.3154767671373695,0.5744374818469364), linewidth(0.7) + uququq); draw(circle((-5.029501701930812,2.1310878719947333), 4.027051235509019), linewidth(0.7) + uququq); draw((-8.743526636724253,0.5744374818469364)--(-5.029501701930812,6.158139107503752), linewidth(0.7) + blue); draw((-5.029501701930812,6.158139107503752)--(-1.3154767671373695,0.5744374818469364), linewidth(0.7) + blue); draw((-8.743526636724253,0.5744374818469364)--(-3.3274325020730524,-1.5185836327177504), linewidth(0.7) + uququq); draw((-3.3274325020730524,-1.5185836327177504)--(-1.3154767671373695,0.5744374818469364), linewidth(0.7) + uququq); draw((-8.743526636724253,0.5744374818469364)--(-4.462145301978226,3.599231527429918), linewidth(0.7) + ffqqff); draw((-4.462145301978226,3.599231527429918)--(-1.3154767671373695,0.5744374818469364), linewidth(0.7) + uququq); draw((-5.029501701930812,6.158139107503752)--(-4.462145301978226,3.599231527429918), linewidth(0.7) + uququq); draw((-4.462145301978226,3.599231527429918)--(-3.3274325020730524,-1.5185836327177504), linewidth(0.7) + ffqqff); draw((-5.029501701930812,0.5744374818469364)--(-5.029501701930812,6.158139107503752), linewidth(0.7) + uququq); /* dots and labels */ dot((-8.743526636724253,0.5744374818469364),linewidth(3pt) + dotstyle); label("$B$", (-9.4,0.38981359230911394), NE * labelscalefactor); dot((-1.3154767671373695,0.5744374818469364),linewidth(3pt) + dotstyle); label("$C$", (-1.1777051057649721,0.371945341379836), NE * labelscalefactor); dot((-3.3274325020730524,-1.5185836327177504),linewidth(3pt) + dotstyle); label("$F$", (-3,-1.7543765192042373), NE * labelscalefactor); dot((-5.029501701930812,6.158139107503752),linewidth(3pt) + dotstyle); label("$A$", (-4.992576679165819,6.214863395253718), NE * labelscalefactor); dot((-4.462145301978226,3.599231527429918),linewidth(3pt) + dotstyle); label("$E$", (-4.429726774893563,3.650769386902336), NE * labelscalefactor); dot((-3.791493390332998,0.5744374818469364),linewidth(3pt) + dotstyle); label("$D$", (-3.7596673650456394,0.631034979854366), NE * labelscalefactor); dot((-5.029501701930812,0.5744374818469364),linewidth(3pt) + dotstyle); label("$M$", (-4.992576679165819,0.631034979854366), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] #1885

posted many times https://artofproblemsolving.com/community/q3h1748556p11390757 https://artofproblemsolving.com/community/q3h1197204p5868890 https://artofproblemsolving.com/community/q3h534552p3063716

A solution, maybe similar to those in other links at https://stanfulger.blogspot.com/2021/09/portugal-2021-aops.html. Best regards, sunken rock