Tried to come up with a non-calculation solution, but alas. Here's a solution that's at least somewhat clean. Let $I$ be the incenter of $\triangle ABC$. Observe that $\triangle ABD\sim\triangle ACI$ (call this $(*)$), so \begin{align*} \frac{BD}{CE} &= \frac{BD}{CI}\cdot \frac{CI}{CE} \stackrel{(*)}=\frac{AB}{AC}\cdot\frac{CI}{CE}\\ &= \frac{\sin 40^\circ}{\sin 20^\circ}\cdot\frac{1}{\cos 20^\circ} = \frac{2\sin 20^\circ \cos 20^\circ}{\sin 20^\circ\cos 20^\circ} = 2. \end{align*}

Non calculation solution: Note that $\angle CDI=\angle DIC=80^\circ$ where $I$ is the incentre so $CD=CI$. Suppose that $(C,I)$ intersects $BC$ at $F\neq D$. Then $\angle BIE=80^\circ$, and we see $\angle EIF=\angle EDI=80^\circ$ by Thales. Yet $E$ is foot from $I$ to $BC$, so $BE=EF$. Note \[EF=CE+CF=CE+CD=BC-CE\]or $BD=BC-CD=2CE$.

Let $J$ be the $A-$excenter of $\triangle ABC$, and let $F$ be the foot of perpendicular from $J$ onto $BC$. Then it is well known that $CE = BF$, so it suffices to prove that $F$ is the midpoint of $BD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.9459439968693775cm); 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 = 39.021835089749345, xmax = 117.94071502713689, ymin = -18.798847784073523, ymax = 17.018759951543764; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw(arc((47.43137111404738,7.3863501660448145),2.2526797317998346,0.6030911943805356,30.60309119438053)--(47.43137111404738,7.3863501660448145)--cycle, linewidth(0.8)); draw(arc((66.2561440085277,7.58450567019724),2.2526797317998346,130.60309119438054,180.60309119438054)--(66.2561440085277,7.58450567019724)--cycle, linewidth(0.8)); draw(arc((47.43137111404738,7.3863501660448145),2.2526797317998346,-74.39690880561953,0.6030911943805374)--(47.43137111404738,7.3863501660448145)--cycle, linewidth(0.8)); draw(arc((58.821660568078975,7.506247949771463),2.2526797317998346,180.6030911943805,260.60309119438057)--(58.821660568078975,7.506247949771463)--cycle, linewidth(0.8)); draw(arc((60.03557625959951,14.841374192106537),2.2526797317998346,-99.39690880561955,-49.396908805619475)--(60.03557625959951,14.841374192106537)--cycle, linewidth(0.8)); /* draw figures */ draw((60.03557625959951,14.841374192106537)--(47.43137111404738,7.3863501660448145), linewidth(0.8) + wrwrwr); draw((47.43137111404738,7.3863501660448145)--(66.2561440085277,7.58450567019724), linewidth(0.8) + wrwrwr); draw((66.2561440085277,7.58450567019724)--(60.03557625959951,14.841374192106537), linewidth(0.8) + wrwrwr); draw(circle((59.35278099729571,10.715560940287288), 3.203544772366925), linewidth(0.8) + wrwrwr); draw((47.43137111404738,7.3863501660448145)--(54.57088665924533,-18.179195830812297), linewidth(0.8) + wrwrwr); draw((60.03557625959951,14.841374192106537)--(54.57088665924533,-18.179195830812297), linewidth(0.8) + wrwrwr); draw((54.57088665924533,-18.179195830812297)--(54.30101446943294,7.458662201364665), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((66.2561440085277,7.58450567019724),linewidth(2.pt) + dotstyle); label("$C$", (67.05399187134736,6.279557373004669), NE * labelscalefactor); dot((47.43137111404738,7.3863501660448145),linewidth(2.pt) + dotstyle); label("$B$", (44.93005293393551,6.280986563297911), NE * labelscalefactor); label("20", (50.435412397535174,7.857862375557792), NE * labelscalefactor); label("40", (61.924079029714335,8.23330899752443), NE * labelscalefactor); dot((60.03557625959951,14.841374192106537),linewidth(2.pt) + dotstyle); label("$A$", (60.34720321745445,15.141526841710574), NE * labelscalefactor); dot((59.386500653142136,7.512193634877391),linewidth(2.pt) + dotstyle); label("$E$", (59.29595267594786,5.830450616937945), NE * labelscalefactor); dot((54.30101446943294,7.458662201364665),linewidth(2.pt) + dotstyle); label("$F$", (54.86568253674152,5.379914670577979), NE * labelscalefactor); dot((54.57088665924533,-18.179195830812297),linewidth(2.pt) + dotstyle); label("$J$", (55.241129158708155,-18.198133188926903), NE * labelscalefactor); dot((58.821660568078975,7.506247949771463),linewidth(2.pt) + dotstyle); label("$D$", (57.3941661880813,8.308755619491069), NE * labelscalefactor); label("80", (49.53434050481524,4.629021426644703), NE * labelscalefactor); label("80", (55.99202240264143,4.553932102251375), NE * labelscalefactor); label("60", (60.12193524427446,11.236881973257535), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now, note that $\angle JBD = (180^\circ - 20^\circ)/2 = 80^\circ$ and $\angle JDB = \angle ADC = 180^\circ - 60^\circ - 40^\circ = 80^\circ$. Hence $JD = JB$ which implies that $F$ is indeed the midpoint of $BD$. $\blacksquare$