Let $T_1=BK\cap DE$ and $T_2=CL\cap DE$ then the radical axes of $(ADE), (ABC)$ and $(BDE)$ are concurrent at $T_1$ so $T_1A$ is tangent to $(ABC)$. Similarly $T_2A$ is tangent to $(ABC)$ therefore $A,T_1$ and $T_2$ are collinear. Meaning $T_2=T_1=T$ and we’re done.

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Note that $T,D,E$ is collinear,$D,L,C,E$is cyclic So $TD\cdot TE=TL\cdot TC$ Which meas that $Pow(T,\odot(ABC))=Pow(T,\odot(ADE))$ Hence $T$ is on the redical axis of $\odot(ABC)$and$\odot(ADE)\Rightarrow$ $TA$ is tangent to the $\odot(ABC)$

My solution from the exam. Let the tangent to $(ABC)$ and $CL$ intersect at $T'$ and $T'B \cap (ABC)=K'$. We'll prove that $B,D,E,K'$ are cyclic. $BD \cap LC=Q$. $\angle DLC=\angle DEA=\angle C=\angle T'AD \implies T',A,D,L$ are cyclic. $\angle LCB=\angle LAB= \angle LT'D \implies T'D \parallel BC \implies T',D,E$ are collinear. $|T'D|.|T'E|=|T'A|^2=|T'B|.|T'K'| \implies E,D,B,K'$ are cyclic.

The radical axes of circles $(B D E)$, $(C D E)$ and $(A B C)$ are concurrent so $DE$, $CL$ and $BK$ are concurrent, let their concurrency point be $T$. The radical axes of circles $(A D E)$, $(A B C)$ and $(B D E)$ are concurrent so $AA$, $DE$ and $BK$ are concurrent, we know that intersection point of $DE$ and $BK$ is $T$, so $DE$, $BK$, $CL$ and $AA$ are concurrent at point $T$, so $TA$ is tangent to $(A B C)$, as desired.

[asy][asy] import graph; size(13cm); 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 = -1.85536646698403, xmax = 35.12511020013301, ymin = -5.306020920464698, ymax = 13.877321872997403; /* image dimensions */ pen qqzzff = rgb(0.,0.6,1.); filldraw((4.,0.)--(11.827055398593837,-3.8519543758649073)--(8.767275410968335,4.579920052837284)--cycle, qqzzff + opacity(0.05000000074505806), linewidth(1.) + qqzzff); filldraw((11.827055398593837,-3.8519543758649073)--(25.3395450740991,4.579920052837284)--(15.62265479609936,11.165880965339193)--cycle, qqzzff + opacity(0.05000000074505806), linewidth(1.) + qqzzff); /* draw figures */ draw((15.62265479609936,11.165880965339193)--(4.,0.), linewidth(1.)); draw((4.,0.)--(18.,0.), linewidth(1.)); draw((18.,0.)--(15.62265479609936,11.165880965339193), linewidth(1.)); draw(circle((11.,4.345641664874962), 8.239211217070311), linewidth(1.)); draw(circle((12.89607872981268,1.3070831003254622), 5.268631564593121), linewidth(1.)); draw((25.3395450740991,4.579920052837284)--(15.62265479609936,11.165880965339193), linewidth(1.)); draw((15.62265479609936,11.165880965339193)--(11.827055398593837,-3.8519543758649073), linewidth(1.)); draw((11.827055398593837,-3.8519543758649073)--(25.3395450740991,4.579920052837284), linewidth(1.)); draw((4.,0.)--(11.827055398593837,-3.8519543758649073), linewidth(1.)); draw((11.827055398593837,-3.8519543758649073)--(8.767275410968335,4.579920052837284), linewidth(1.)); draw((4.,0.)--(11.827055398593837,-3.8519543758649073), linewidth(1.) + qqzzff); draw((11.827055398593837,-3.8519543758649073)--(8.767275410968335,4.579920052837284), linewidth(1.) + qqzzff); draw((8.767275410968335,4.579920052837284)--(4.,0.), linewidth(1.) + qqzzff); draw((11.827055398593837,-3.8519543758649073)--(25.3395450740991,4.579920052837284), linewidth(1.) + qqzzff); draw((25.3395450740991,4.579920052837284)--(15.62265479609936,11.165880965339193), linewidth(1.) + qqzzff); draw((15.62265479609936,11.165880965339193)--(11.827055398593837,-3.8519543758649073), linewidth(1.) + qqzzff); draw((8.767275410968335,4.579920052837284)--(25.3395450740991,4.579920052837284), linewidth(1.)); /* dots and labels */ dot((15.62265479609936,11.165880965339193),dotstyle); label("$A$", (15.71818532283096,11.417918950758672), NE * labelscalefactor); dot((4.,0.),dotstyle); label("$B$", (3.2211707116372647,-0.4542897011392019), NE * labelscalefactor); dot((18.,0.),dotstyle); label("$C$", (18.199946453453688,-0.4542897011392019), NE * labelscalefactor); dot((8.767275410968335,4.579920052837284),dotstyle); label("$D$", (8.407051181266707,4.866963894250053), NE * labelscalefactor); dot((17.024882048657027,4.579920052837285),linewidth(4.pt) + dotstyle); label("$E$", (17.10439424263825,4.7551728523301104), NE * labelscalefactor); dot((11.827055398593837,-3.8519543758649073),linewidth(4.pt) + dotstyle); label("$L$", (11.693707813713024,-4.58050793489148), NE * labelscalefactor); dot((25.3395450740991,4.579920052837284),linewidth(4.pt) + dotstyle); label("$T$", (25.421647761481985,4.7551728523301104), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] By radical axis theorem $CL$, $BK$, $DE$ are concurrent at $T$. Observe that there is a spiral similarity centered at $L$ which maps $\triangle LDT$ to $\triangle LBA$, because $\angle ALD = \angle ACB = \angle AED = \angle DLC = \angle DLT$ and $\angle BAL = \angle BCL = \angle DTL $. Thus by the properties of spiral similarity we get that $\triangle LAT \thicksim \triangle LBD $, so $\angle LAT = \angle LBD = \angle LBA $. $\square$