Nice problem! With an easy angle chasing you get $\angle LKA=\angle LCB$ then $BLCK$ is cyclic. Inversion centered at $A$ with radius $\sqrt{AB\cdot AK}$. Line $BC$ is the image of $(ADE)$ and line $KL$ is the image of $ABC$ then the intersection $P=BC\cap KL$ is the image of $M=(ADE)\cap (ADE)$, which implies $A,P,M$ are collinear and we are done.$\Box$

Any elementary solution??

ayan_mathematics_king wrote: Any elementary solution?? From #2 we get $CLBK$ is cyclic. Hence, $AM$ is the radical axis of $\odot(ABC)$ and $\odot(ADE)$, $KL$ is the radical axis of $\odot(BKCL)$ and $\odot(ADE)$ and $BC$ is the radical axis of $\odot(BKCL)$ and $\odot(ABC)$. Hence, $AM,BC,KL$ concur.

I solved like post no.4 Post no.2 is more creative YEAH!!!

$\angle BKL = \angle AKL = \angle ADL = \angle AOD = \angle ACB = \angle BCL$ so $BKCL$ is cyclic. Let $KL$ meet $BC$ at $M$. $BM.MC = KM.ML$ so $M$ has same power w.r.t both $ADE$ and $ABC$ so $M$ lies on radical axis of them so $M$ lies on $AM$.

Why DOL is collinear?