Let $FT \cap BN=X$ and $ET \cap MC=Y$ From $T$ is miquel point it is easy to show $X-A-Y$ and $(B,T,A,X)$,$(A,T,Y,C)$,$(X,Y,B,C)$. Finally look at the radical axises of these three cyclic quadrilateral.

Let $MN\cap AK=T$ $K$ s the miquel point of $AEIF$. \[\angle BKT=\angle BKF+\angle FKT=\angle BIF+\angle FCA=\frac{\angle B}{2}+\angle C\]\[\angle TNB=\angle MNB=\angle MNC+\angle CNB=\angle A+\frac{\angle B}{2}\]So $NTKB$ is cyclc. Similarily, $TKMC$ is cyclic. Radical axises of $(BNMC),(BNTK),(MCTK)$ are concurrent which gives that $BN, MC, AK$ are concurrent as desired.$\blacksquare$

Let $NB \cap MC=X$.If $X,A,K$ are collinear,we're done. Claim:$XBKC$ is cyclic. $\angle BXC=180-3(\angle FBI +\angle ECI)=180- 3\angle FKE=180-\angle BKC$. So we need to prove $\angle XBC=\angle AKC \implies AEKB$ should be cyclic,but $\angle KEC=\angle KIC=\angle FBK=\angle ABK$.

By $\sqrt{bc}-$inversion we have that $\mathbb{I} (B)=C$ Let $I_{a}$ the $A$-excenter of triangle $ABC$, we have that $\mathbb{I} (I)=I_{a}$ We know that $AE=\frac{bc}{a+c}$ and $E$ is on the ray $AC$, then $\mathbb{I} (E)=P$, so that $AP=a+c$ and $P$ is on the ray $AB$ Similarly, $\mathbb{I} (F)=Q$, so that $AQ=a+b$ and $Q$ is on the ray $AC$ $\Rightarrow PB=BC=CQ=a$ $\Rightarrow PC \perp BI_a$ and $BQ \perp CI_a$ Let $R \equiv PC \cap BQ$, we will prove that $\mathbb{I} (K)=R$ $PC \perp BI_a$ and $BQ \perp CI_a \Rightarrow R$ is orthocenter of $BI_aC$ $\Rightarrow \angle BCR = \angle BI_aR$, also $\angle BCP = \angle BPC \Rightarrow BPI_aR$ is cyclic Similarly, $CQI_aR$ is cyclic $\Rightarrow R \equiv (BPI_a) \cap (CQI_a) \Rightarrow R = \mathbb{I} (K)$ We know that $BI \perp BI_a$ and $CI \perp CI_a$ $\Rightarrow BICR$ is parallelogram $M$ is the midpoint of the arc $AC$ that does not contain $B \Rightarrow \mathbb{I} (M) = U$, so that $U$ is in the extension of the line $CB$ through $B$ and $UB=AB$ Similarly, $\mathbb{I} (N) = V$, so that $V$ is in the extension of the line $BC$ through $C$ and $VC=AC$ Let $X \equiv BN \cap CM \Rightarrow \mathbb{I} (X) \equiv (ACV) \cap (ABU)$ $\Rightarrow AX$ is the radical axis of $(ACV)$ and $(ABU)$ Let $Y \equiv AR \cap BC$, it is enough to show that $Y$ is on the radical axis of $(ACV)$ and $(ABU)$ $\Leftrightarrow YB \times YU=YC \times YV \Leftrightarrow \frac{YB}{YC}=\frac{a+b}{a+c} \Leftrightarrow \frac{[ABP]}{[ACP]}=\frac{a+b}{a+c}$ The latter is verified since $BICP$ is parallelogram (just plot the heights and use the midpoint of $IP$)

A lot easier with radical axes but another solution is another solution Let $BN \cap CM = Z$. We already know $(AEKB)$ and $(AFKC)$ and we are going to prove $Z$ is on $AK$. İf we prove $Z$ has same power for $(AEKB)$ and $(AFKC)$ the proof ends. So we want to prove $ZB.ZX = ZC.ZY$ We also know $ZB.ZN = ZC.ZM$. Hence, it suffices to show that $MN || XY$. $\angle BNM = \angle A + \angle \frac{B}{2}$ $\angle BXA = \angle BEC = \angle A + \angle \frac{B}{2}$ hence it suffices to show $M$, $A$ and $N$ are collinear. We are going to show $\angle ZXA + \angle ZYA + \angle XZY = 180$ and this ends the proof. $\angle ZXA = \angle A + \angle \frac{B}{2}$ and similarly $\angle ZYA = \angle A + \angle \frac{C}{2}$ $\angle XZY = 180 - \angle ZMN - \angle ZNM = 180 - 2.\angle A - \angle \frac{B}{2} - \angle \frac{C}{2}$ and we are done. $\blacksquare$