We need to show $\triangle CEL$ is isosceles or $CL=LE$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6.cm); 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 = -7.06, xmax = 4.54, ymin = -3.08, ymax = 6.3; /* image dimensions */ pair A = (0.,4.82), B = (-2.,0.), C = (4.,0.), L = (0.7269362485907217,0.), D = (0.36346812429536085,2.41), F = (1.332579449351194,3.214241763531812), G = (2.665158898702388,1.608483527063623); draw(A--B--C--cycle, linewidth(1.)); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(A--L, linewidth(1.)); draw((-0.79972854037171,2.892654217704179)--D, linewidth(1.)); draw(D--F, linewidth(1.)); draw(L--G, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (0.08,5.02), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-2.32,-0.3), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (4.08,0.2), NE * labelscalefactor); dot(L,linewidth(4.pt) + dotstyle); label("$L$", (0.58,-0.4), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (0.04,1.98), NE * labelscalefactor); dot((-0.79972854037171,2.892654217704179),linewidth(4.pt) + dotstyle); label("$E$", (-1.,3.06), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (1.42,3.38), NE * labelscalefactor); dot(G,linewidth(4.pt) + dotstyle); label("$G$", (2.74,1.76), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $F$ and $G$ be points on $AC$ such that $DF\perp AC$ and $LG \perp AC$. As $AD$ is angle bisector of $\angle EAF$ and $DE\perp AE;DF\perp AF$, we have $\triangle AED \cong \triangle AFD \implies AE=AF$. So, from midpoint theorem in $\triangle ALG$ and as $AD=DL$, we have $2AE=2AF=AG$. From the condition we have, \[AC=3AE=2AE+AE=AG+AE=AG+AF.\]So, $AF = AC - AG = GC\Rightarrow AF=FG=GC$. Lets consider $\triangle LFC$. We have $LG\perp FC$ and $FG=GC$ which gives $LF=LC$. Consider $\triangle LDF$ and $\triangle LDE$. We have $ED=DF$ and $\angle ADE=\angle ADF$ which gives $\triangle LDF\cong \triangle LDE$. So, $LE=LF$. So we have $LC=LF=LE$ which gives $\triangle CEL$ is isosceles. Proved.

Simple but nice problem! TheDarkPrince wrote: Let $AL$ be the bisector of triangle $ABC$, $D$ be its midpoint, and $E$ be the projection of $D$ to $AB$. It is known that $AC = 3AE$. Prove that $CEL$ is an isosceles triangle. Solution. Let $AL=y $ and $ AE=x$. Claim: $ EL = CL $ Proof $ i) $ By Cosine Rule in $\triangle AEL $ we have $ EL^2 = x^2 + y^2 - 2xy cos\alpha $ where $\alpha = \dfrac{A}{2} .$ $ii)$ By Cosine Rule in $\triangle CAL $ we have $ CL^2 = y^2 + 9x^2 - 6xy cos\alpha $ Now we show that both these equations are equal. Therefore $ EL=CL \iff x^2 = 9x^2-4xy cos\alpha \iff y cos\alpha = 2x .$ But this is true by Sine Rule in $\triangle ADE$ $ \dfrac{ AE}{sin(90 - \alpha)} = \dfrac { AE}{\cos \alpha } = AD= \dfrac{y}{2} $ QED.

see the fgure: