We can find $AE^2$ from triangle $AOE$ $AE^2=2AO^2 (1+\cos \angle BAC)= \frac{k^2AB^2}{2} (1+\cos \angle BAC)$ And $AD^2= AC *AB (1 - \frac{CB^2}{(AC+AB)^2})= \frac{2 AC^2*AB^2}{(AC+AB)^2}(1+\cos \angle BAC)=\frac{2k^2AB^2}{(k+1)^2}(1+\cos \angle BAC)$ So $\frac{AD}{AE}=\frac{2}{k+1}$

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.760433772373926, xmax = 23.920873074754375, ymin = -26.36714817636629, ymax = 11.432112570887025; /* image dimensions */ /* draw figures */draw((-5.8241148128521365,4.93287259008292)--(-8.216473087994752,-8.504206388634767), linewidth(0.8)); draw((-8.216473087994752,-8.504206388634767)--(7.812327355460768,-8.42446111279668), linewidth(0.8)); draw((-0.678858311163229,-11.14227866621626)--(-5.8241148128521365,4.93287259008292), linewidth(0.8)); draw((-5.8241148128521365,4.93287259008292)--(7.812327355460768,-8.42446111279668), linewidth(0.8)); draw((7.812327355460768,-8.42446111279668)--(-0.678858311163229,-11.14227866621626), linewidth(0.8)); /* dots and labels */dot((-5.8241148128521365,4.93287259008292),dotstyle); label("$A$", (-5.664624261175961,5.331598969273355), NE * labelscalefactor); dot((-8.216473087994752,-8.504206388634767),dotstyle); label("$B$", (-8.056982536318575,-8.105480009444332), NE * labelscalefactor); dot((7.812327355460768,-8.42446111279668),dotstyle); label("$C$", (7.971817907136941,-8.025734733606244), NE * labelscalefactor); dot((-1.5338811506011178,-8.47095966257808),linewidth(4pt) + dotstyle); label("$D$", (-1.3583793659192551,-8.145352647363374), NE * labelscalefactor); dot((-0.678858311163229,-11.14227866621626),linewidth(4pt) + dotstyle); label("$E$", (-0.5210539696193401,-10.816819387939294), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] We have $$\frac{AD}{AC}=\frac{\sin C}{\sin(B+\frac{A}{2})}$$and $$\frac{AE}{AC}=\cos\frac{A}{2}$$Now by product to sum formula $$\frac{AD}{AE}=\frac{\sin C}{\sin(B+\frac{A}{2})\cos\frac{A}{2}}=\frac{2\sin C}{\sin C+\sin B}=\frac{1}{\frac{1}{2}+\frac{k}{2}}=\frac{2}{k+1}$$as desired.

[asy][asy] size(7cm); import geometry; import olympiad; pen b = rgb(20,105,176); point O = (0,0); point A = O+2*dir(130); point B = O+2*dir(210); point C = O+2*dir(330); point M = (0,-2); path AM = A -- M; path BC = B -- C; pair D = intersectionpoints(AM,BC)[0]; pair F = foot(B,A,M); pair E = foot(C,A,M); draw(circle(O,2),gray); draw(A--B--C--cycle,b); draw(A--B--F--cycle,green); draw(A--C--E--cycle,green); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); label("$A$",A,NW); label("$B$",B,W); label("$C$",C,E); label("$D$",D,SW); label("$E$",E,S); label("$F$",F,NE); [/asy][/asy] We construct the point $F$ on $AD$ such that $BF\perp AD$, so we have the two pairs of similar triangles \[\triangle ABF\sim\triangle ACE~~~~~\text{and}~~~~~\triangle DBF\sim \triangle DCE.\] This gives the ratios \[\dfrac{AF}{AE}=\dfrac{AB}{AC}=\dfrac{1}{k}~~~~~\text{and}~~~~~\dfrac{DF}{DE}=\dfrac{DB}{DC}=\dfrac{1}{k},\](which implies that $(A,D;E,F)=-1$, as a note) and so we can write $AF=x,AE=xk,DF=y$ and $DE=yk$ for some $x$ and $y$ such that $AF+FD+DE=AE$, i.e. \[x+y+yk=xk\Longrightarrow 1+\dfrac{y}{x}+\dfrac{y}{x}k=k\Longrightarrow \dfrac{y}{x}=\dfrac{k-1}{k+1}.\] From here, our desired ratio is hence \[\dfrac{AD}{AE}=\dfrac{x+y}{xk}=\dfrac{1}{k}+\dfrac{y}{xk}=\dfrac{1}{k}\left(1+\dfrac{k-1}{k+1}\right)=\dfrac{2}{k+1}.\]

You can also extend $CE$ to cut $AB$ at $F$, then apply Menelaus' Theorem in $\triangle DEC$ with $\overline{ABF}$ to get the desired result.

let $CE$ cuts $AB$ at $C'$ and the parallel to $CE$ through $D $ cuts $AB$ at $G$ then $\frac{AD}{AE}=\frac{GD}{C'E}=2.\frac{GD}{C'C}=2.\frac{BD}{BC}=2.\frac{AB}{AB+AC}=2.\frac{1}{1+k}$

Dear PROF65, very elegant proof.. Sincerely Jean-Louis