First , note that while we have $\triangle ALC \sim \triangle ABK$ and also since $EB \parallel FC$ , we have $\triangle AEB \sim \triangle ACF$ and we can get : $$AL.AK=AB.AC=AE.AF$$Thus $\triangle ALF \sim \triangle AEK$ and if point $S$ be the intersection point of lines $EK$ and $FL$ , then quadrilaterals $AESL$ and $ASFK$ are cyclic and $\angle LSK=90+\frac{\angle A}{2}$. Now let point $T$ be the second intersection point of circumcircles of triangles $\triangle FIL$ and $\triangle EIK$ , then one can see that : $$\angle FTE=\angle FTI-\angle ETI=180-\angle FLI-\angle EKI=180-(\angle SLI+\angle SKI)=\angle A$$Thus the quadrilateral $AEFT$ is cyclic and we're done.

denote $X$ as the intersection point of circumcircles $FIL ,EIK$ and prove $XEAF$ is cyclic quadrilateral by proving $$\angle EXF=180-\angle A$$$\angle{EXI}= \angle{EKI}$ $\angle{IXF}= \angle{LFI}$ so we need to prove $\angle EKI+\angle FLI=\angle B+\angle C$ and you can use the fact that $\triangle ALF \sim \triangle AEK$

Let circles $(FIL)$ and $(EIK)$ intersect at $P\neq I$. Then $\angle EPF=\angle EPI+\angle FPI=\angle EKI+\angle FLI=\angle ALF+\angle AKE+\frac{\angle B+\angle C}{2}$. However notice that triangle $ALF$ and $AEK$ are similar as $\angle LAF=\angle EAK$ and $EA\cdot AF=AB\cdot AC=AL\cdot AK$. Then we get $\angle EPF=(180-\angle LAF)+\frac{\angle B+\angle C}{2}=180-\angle A$. This implies that P lies on $(AEF)$ and finishes the proof.

matinyousefi wrote: In a triangle $ABC$ the incenter, the $B$-excenter and the $C$-excenter are $I, K$ and $L$, respectively. The perpendiculars at $B$ and $C$ to $BC$ intersect the lines $AC$ and $AB$ at $E$ and $F$, respectively. Prove that the circumcircles of $AEF, FIL, EIK$ concur. Let $S$ is an insection of circles $(ILF)$ and $(IKE)$ ($S \neq I$) Cuz $BE \| CF$ so $\frac{AE}{AC} = \frac{AB}{AF} $ (1) We already know that $A$, $I$, $K$, $C$ are concylic; $A$, $I$, $L$, $B$ are concylic Therefore, $\angle ALB = \angle AIK = \angle ACK$, $\angle ABL = \angle AKC$. Which means $\triangle ALB \sim \triangle ACK$ That lead to $\frac{AL}{AC} = \frac{AB}{AK}$ (2) From (1), (2), we see that $\frac{AL}{AF} = \frac{AE}{AK}$. So $\triangle ALE \sim \triangle AFK$ (Cuz $LAK$ is the exterior angle bisector of $\angle BAC$ Thus, $\angle ESF = \angle ISE + \angle ISF = \angle IKE + \angle ILF = (\angle AKE + \frac{1}{2}\angle ACB) + (\angle ALF + \frac{1}{2}\angle ABC)$ $\hspace{2.1cm} = \angle AKE + \angle AEK + \frac{1}{2}\angle ACB + \frac{1}{2}\angle ABC) = \angle EAL + \frac{1}{2}\angle ACB + \frac{1}{2}\angle ABC) = 180^\circ - \angle EAF $ Therefore, $S$ lie on circe $(AEF)$, done

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