Lemma. Let the feet from $A$, $B$, and $C$ in $\bigtriangleup ABC$ be $D$, $E$, and $F$. $H$ is the orthocenter and $M$ and $N$ are the midpoints of $BC$ and $EF$, respectively. Then $\bigtriangleup BHF\sim \bigtriangleup MFN\sim \bigtriangleup MEN\sim \bigtriangleup CHE$. Proof. Angle chase. Back to main problem Let $l_E$ and $l_F$ intersect at $K$ and redefine $D$ as the foot from $K$ to $BC$. $X=EK\cap AB$, $Y=FK\cap AC$, $S,N,Z$ are the midpoints of $AK$, $EF$, $XY$, respectively. It suffices to show $D\in(EFM)$. By Newton gauss, points $S$, $N$, and $Z$ are colinear. Note that the circles $(AEKF)$, $(M,ME)$ share $EF$ as common chord, so $SM$ bisects this chord, i.e. $S$, $N$, and $M$ are colinear. In particular, points $N$, $Z$, and $M$ are colinaer. Now by converse ERIQ lemma, we get $\frac{FX}{XB}=\frac{NZ}{ZM}=\frac{EY}{YC}$. Combine this with lemma and we have $\bigtriangleup BKF\sim\bigtriangleup MFN\sim \bigtriangleup MEN\sim \bigtriangleup CKE$. To finish, note that $KFBD$ and $KECD$ are cyclic, so $$\angle EDF=\angle EDK+\angle KDF=\angle ECK+\angle KBF=\angle EMN+\angle FMN=\angle EMF$$

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Let the perpenediculars from $B$ to $FD$ and from $C$ to $ED$ intersect $FD$ at $P$ and $ED$ at $Q$, respectively, and intersect each other at $X$. Claim: Points $P,Q,X,D,M$ are concyclic. Proof: Note that $\angle XBC=90^\circ-\angle FDB=90^\circ-\angle FEM=90^\circ-\angle EFM=90^\circ-\angle EDM=\angle XCB,$ and so triangle $XBC$ is isosceles, hence $\angle XMD=90^\circ,$ and so all five points $P,Q,X,D,M$ are concyclic on a circle of diameter $XD$ $\blacksquare$ Claim 2: $FP=EQ$. Proof: Note that triangles $FMP$ and $EMQ$ are equal, since $\angle MPQ=\angle QDM=\angle EDM=\angle FEM=\angle PQM,$ and so $MP=MQ$. Moreover, $FM=ME$, and lastly $\angle QMP=\angle FDE=\angle FME$. Thus, these two triangles are equal, as desired. Hence, we infer that $FP=EQ$ $\blacksquare$ Now, let $\ell_D$ and $\ell_F$ intersect at point $Y$, and $\ell_D$ and $\ell_E$ intersect at point $Z$. We need to prove that $Y \equiv Z$, that is $DY=DZ$. Note that $DY=BY\cos \angle BYD=BY\cos \angle BFD=BY \cdot \dfrac{FP}{FB}=\dfrac{FP}{\sin \angle FYB}=\dfrac{FP}{\sin \angle FDB},$ and similarly $DZ=\dfrac{EQ}{\sin \angle EDC}$. However, the two ratios are equal, since $FP=EQ$ from our Claim, and $\angle FDB=\angle FEM=\angle EFM=\angle EDC$. Hence, $DY=DZ$, as we aimed to prove.

Note that D is the unique point on line BC such that BC is the external bisector of the angle between the lines formed between that point and E, F. It suffices to show that D is the projection of the antipode A’ of A with respect to (AEF) onto BC, so by angle chasing it suffices to show that <A’BF = <A’CE, which is equivalent to BF / CE = FA’ / EA’. Note that if B’, C’ are the projections of B, C onto line EF then this simplifies to BF / CE = sin <ECC’ / sin <FBB’, or B’F = C’E. However, this is obvious because the midpoint of E’F’ is the projection of M onto EF, which is equidistant from E and F, so we are done.

suppose $\ell_D, \ell_E$ insect at J， and J' is on $\ell_D$ to use the sin form of Ceva's theorem ，we only need to prove $\begin{aligned} & \frac{\sin \angle \mathrm{FEJ}}{\sin \angle \mathrm{DEJ}} \cdot \frac{\sin \angle \mathrm{EDJ'}}{\sin \angle \mathrm{FDJ'}} \cdot \frac{\sin \angle \mathrm{DFJ}}{\sin \angle \mathrm{EFJ}}=1 \\ & \text { Obviously, } \angle \mathrm{EDJ'}=\angle \mathrm{FDJ'} \\ & \text { We only need to prove that } \\ & \frac{\sin \angle \mathrm{FEJ}}{\sin \angle \mathrm{DEJ}}=\frac{\sin \angle \mathrm{EFJ}}{\sin \angle \mathrm{DFJ}} \\ & \text { Since } \angle \mathrm{FEJ}=\angle \mathrm{CEL}=\angle \mathrm{BKF}, \angle \mathrm{EFJ}=\angle \mathrm{BFK} \\ & \text { Therefore, we only need to prove that } \frac{\sin \angle \mathrm{DEJ}}{\sin \angle \mathrm{DFJ}}=\frac{\mathrm{BF}}{\mathrm{CE}}\end{aligned}$ which means $BF \cdot cos\angle BFD=CE \cdot cos\angle CED$ and the rest is same as @Orestis_Lignos done. Orestis_Lignos wrote: Let the perpenediculars from $B$ to $FD$ and from $C$ to $ED$ intersect $FD$ at $P$ and $ED$ at $Q$, respectively, and intersect each other at $X$. Claim: Points $P,Q,X,D,M$ are concyclic. Proof: Note that $\angle XBC=90^\circ-\angle FDB=90^\circ-\angle FEM=90^\circ-\angle EFM=90^\circ-\angle EDM=\angle XCB,$ and so triangle $XBC$ is isosceles, hence $\angle XMD=90^\circ,$ and so all five points $P,Q,X,D,M$ are concyclic on a circle of diameter $XD$ $\blacksquare$ Claim 2: $FP=EQ$. Proof: Note that triangles $FMP$ and $EMQ$ are equal, since $\angle MPQ=\angle QDM=\angle EDM=\angle FEM=\angle PQM,$ and so $MP=MQ$. Moreover, $FM=ME$, and lastly $\angle QMP=\angle FDE=\angle FME$. Thus, these two triangles are equal, as desired. Hence, we infer that $FP=EQ$ $\blacksquare$ Now, let $\ell_D$ and $\ell_F$ intersect at point $Y$, and $\ell_D$ and $\ell_E$ intersect at point $Z$. We need to prove that $Y \equiv Z$, that is $DY=DZ$. Note that $DY=BY\cos \angle BYD=BY\cos \angle BFD=BY \cdot \dfrac{FP}{FB}=\dfrac{FP}{\sin \angle FYB}=\dfrac{FP}{\sin \angle FDB},$ and similarly $DZ=\dfrac{EQ}{\sin \angle EDC}$. However, the two ratios are equal, since $FP=EQ$ from our Claim, and $\angle FDB=\angle FEM=\angle EFM=\angle EDC$. Hence, $DY=DZ$, as we aimed to prove.

Let $l_E \cap l_F=P$, let $D$ be the foot of the altitude of $P$ on $BC$. $K$ and $L$ are the projections of $B,C$ on $EF$. $H$ is the projection of $A$ onto $EF$. We have $FK=EL$ which gives $$\frac{PF}{PE}\frac{AE}{AF} \overset{\text{ratio lemma}}=\frac{HE}{HF} = \frac{HE}{EL} \frac{FK}{HF} \overset{\text{similarities}}=\frac{EA}{EC}\frac{FB}{FA} \implies \frac{FB}{FP}=\frac{EC}{EP} \overset{\triangle BFP \sim \triangle CEP} \implies \angle FBP=\angle PCE \implies $$ $DP$ is the interior angle bisector of $FDE \implies DM $ is the exterior angle bisector of angle $BDC$. With the fact that $ME=MF$ we get $M \in (DEF)$.

Rephrasing with respect to triangle $DEF$ and changing point names: Rephrased problem wrote: Given a triangle $ABC$, let the exterior bisector of $\angle BAC$ intersect $(ABC)$ again at $D$. Points $E,F$ are taken on line $AD$ such that $DE = FD$. Prove that $(AEB)$ and $(AFC)$ meet on the interior bisector of $\angle BAC$. Let $\ell$ denote the interior bisector of $\angle BAC$. Notice the maps $f : E \mapsto \ell \cap (AEB)$ and $g : E \mapsto F \mapsto \ell \cap (AFC)$ are projective, so it suffices to check $3$ points for $f \equiv g$. The cases $E = I_B$ , $E = I_C$ and $E = D$ are trivial to check, so we are done.

Let $(BFD)$ and $(CED)$ center at $O_B$ and $O_C$ and meet for the second time at a point $S$. Let $P,Q$ are midpoints of $BD,CD$, repsectively. Let $X,Y$ be feet of altitude drawn from $B,C$ to $DF,DE$, respectively. Let $BX,CY$ meet at $T$. Notice that $\angle XBD=90^\circ-\angle FDB=90^\circ-\angle FEM=90^\circ-\angle EFM=90^\circ-\angle CDE=\angle YCD$. Hence, $TB=TC$ and then $TM\perp BC$. This implies that $T,M,D,X,Y$ all lie on the same circle (dashed circle). Since $\angle XDM=\angle YDM$, we have $\angle XYM=\angle YXM$ and then $MX=MY$. Therefore, $\triangle FXM\cong\triangle EYM$ and we now have $FX=EY$. Use angle chasing, one can show that $\triangle FBO_B\sim\triangle ECO_C, \triangle FXB\sim\triangle O_BPB$ and $\triangle EYC\sim\triangle O_CQC$. Consider $$ \frac{O_BP}{O_CQ}=\frac{O_BP}{O_BB}\cdot\frac{O_CC}{O_CQ}\cdot\frac{O_BB}{O_CC}=\frac{FX}{FB}\cdot\frac{EC}{EY}\cdot\frac{BF}{CE}=1 $$Hence, $O_BP=O_CQ$ and then $O_BO_C\parallel BC$. Since $SD$ is a common chord of $(BFD)$ and $(CED)$, then $SD\perp O_BO_C$ which implies that $SD\perp BC$. Since $S$ lies on both $(BFD)$ and $(CED)$, we have $SF\perp AB$ and $SE\perp AC$. Therefore, $l_D,l_E$ and $l_F$ meet at $S$ and we are done.

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A really nice solution by my friend Sepehr Alipour Let $\ell_E$ and $\ell_F$ meet at $P$. Let $D'$ be on $BC$ such that $PD' \perp BC$. Let $F'$ be reflection of $F$ across $M$. Let $\omega$ be circle with center $M$ and radius $ME$. Claim $: EPF$ and $ECF'$ are similar. Proof $:$ First note that since $CF' || AB$ and $\ell_F \perp AB$, we have that $FP$ and $CF'$ meet on $\omega$ so $\angle EFP = \angle EF'C$. we also have $\angle EPF = \angle 180 - \angle A = \angle B + \angle C = \angle ECF'$. Note that now we also have $EPC$ and $EFF'$ are similar so $\angle ED'M = \angle ED'C = \angle EPC = \angle EFF' = \angle EFM$ so $EFD'M$ is cyclic so $D'$ is $D$ and we're Done.

Super short solution: Let $N$ be the point diametrically opposite $M$ on $(MEF)$, then $ND\perp BC$. Let $l_E,l_F$ meet segment $ND$ at $T_E,T_F$. We show $T_E=T_F$. We have $\angle NFM=\angle T_FFC=90,\angle FNM=\angle FDM$, so $F$ spirals $NM$ to $T_FC$. Thus it spirals $NT$ to $MC$. Then $\frac{NT_F}{MC}=\frac{FN}{FM}$. Similarly $\frac{NT_E}{MB}=\frac{EN}{EM}$. But all the lengths are the same so $T_E=T_F$.