I think that we could reduce the problem to the following: Given triangle $ ABC$ and $ (I)$ is its incircle. Let $ A', B'$ be points on $ BC, CA$ and $ d$ is a tangent of $ (I)$. Tangents of $ (I)$ from $ A', B'$ (different from $ BC, CA$) intersect $ d$ at $ A_1, B_1$ respectively. Then $ AA_1, BB_1, A'B'$ are concurrent.

Lets incircle tachs $ AB$, $ BC$, $ CA$ in $ F$, $ E$, $ D$ respectively. Points $ F'$, $ E'$, $ D'$ lie on incircle. $ D' \in A_{1}A'$, $ E' \in B_{1}B'$, $ F' \in C_{1}C'$. $ X = EF \cap LD'$ $ Y = DF \cap LE'$ $ Z = DE \cap LF'$ Let $ P$ be a pole of line $ l$. Then $ P = DD' \cap EE' \cap FF'$ $ AA_{1}$, $ BB_{1}$, $ CC_{1}$ are concurent. $ \Longleftrightarrow$ $ X$, $ Y$, $ Z$ are colinear. ($ X$ is a pole of $ AA_{1}$) From Pascal teorem for $ DEFF'LD'$ we get that $ X$, $ Y$, $ P$ are colinear. Similarly $ X$, $ Z$, $ P$ are colinear. So $ X$, $ Y$, $ Z$, $ P$ are colinear.

My solution is similar to yours. Let $ D, E$ be intersections of $ B'B_1, AB$ and $ l, AC$. We easily note that the hexagon $ B'EA_1A'BD$ circumscribes the incircle of triangle $ ABC$. It follows from Brianchon's Theorem that $ A'B', BE, A_1D$ are concurrent at $ K$. Hence, applying Pappus' Theorem to the 2 triples $ (B_1, A_1, E)$ and $ (A, B, D)$, we have the 3 points $ B', K$ and the intersection $ H$ of $ AA_1, BB_1$ are collinear. Hence, $ H$ is on $ A'B'$. This means $ AA_1, BB_1, A'B'$ are concurrent. The problem then follows from this.

My solution: From Brianchon theorem ( for $ A'CAC'C_1A_1 $ and $ B'ABA'A_1B_1 $ ) we get $ AA_1, BB_1, CC_1, l $ are concurrent . Q.E.D

Before proving the statement I need a prerequisite lemma: Lemma:With respect to a given circle $\omega$,the polars are concurrent if and only if their respective poles are collinear. Proof:Consider the intersection points of the polars and use La Hire's theorem. Now back to our problem.I shall first name the points:$D,E,F$ are the tangency points of the incircle with $BC,CA,AB$ respectively.Let $X$ be the point on $(I)$ such that $B'X$ is tangent to $(I)$ (other than $E$),$Y$ be the point where $l'$ touches $(I)$,$Z$ be the point where $A'A_1$ touches $(I)$ and $L$ be the point where $C'C_1$ touches $(I)$.In the rest of my proof the poles and polars,etc will be considered with respect to the incircle. Note that the polars of $A',B',C'$ are $DZ,EX,FL$ and as $A',B',C'$ are collinear,by our lemma $DZ,EX,FL$ concur.Set $YX \cap DF=P,EF \cap YZ=Q,YL \cap DE=R$.Applying Pascal's theorem on the cyclic hexagon $XYZDFE$ we see that $P,Q,EX \cap DZ$ are collinear.Similarly Pascal's theorem in $XYLFDE$ yeilds that $P,R,XE \cap LF$ are collinear.But $XE \cap DZ=XE \cap LF$ as $DZ,EX,FL$ concur so $P,Q,R$ are collinear. Now see that $DF$ is the polar of $B$,while $XY$ is the polar of $B_1$.Hence by La Hire's theorem $P=DF \cap XY$ is the pole of $BB_1$.Similarly $Q$ is the pole of $AA_1$ and $R$ is the pole of $CC_1$.Finally by our lemma,as $P,Q,R$ are collinear,$AA_1,BB_1,CC_1$ concur,as desired.

Wonderful Problem!!! I guess there is a solution with Dual of Desargues Involution too, can anyone provide that solution too. Iran TST 2008 P2 wrote: Suppose that $ I$ is incenter of triangle $ ABC$ and $ l'$ is a line tangent to the incircle. Let $ l$ be another line such that intersects $ AB,AC,BC$ respectively at $ C',B',A'$. We draw a tangent from $ A'$ to the incircle other than $ BC$, and this line intersects with $ l'$ at $ A_1$. $ B_1,C_1$ are similarly defined. Prove that $ AA_1,BB_1,CC_1$ are concurrent. Let the Tangency Points made by $\odot(I)$ with $\{BC,C,A,AB\}$ be $\{D,E,F\}$ respectively. Let $\ell'$ be tangent to $\odot(I)$ at a point $X$ andlet $\{A'P,B'Q,C'R\}$ be the other tangents where $\{P,Q,R\}\in\odot(I)$. Notice that the Polars of $\{A',B',C'\}$ are $\{PD,QE,RF\}$ respectively WRT $\odot(I)$ and as $\overline{A'-B'-C'}$. Hence, $PD,QE,RF$ are concurrent. Now by Pascal on $XQPDFR$ we get that $XQ\cap FD,QP\cap FR,PD\cap RX$ are collinear. Now Pascal on $XPDEFR$ we get that $XP\cap EF,PD\cap FR,DE\cap RX$ are collinear. So combining both the Collinearities we get that $XP\cap EF, DE\cap XR,FD\cap XQ$ are collinear. Now the Polar of $XP\cap EF$ is $AA_1$, the Polar of $XQ\cap FD$ is $BB_1$ and the Polar of $DE\cap RX$ is $CC_1$. So, $AA_1,BB_1,CC_1$ are concurrent.

Suppose that $\ell$ touches the incircle at $P$. We will prove that $AA_1,B_1,CC_1,\ell$ are concurrent. Firstly, we will show that $BB_1,CC_1,\ell$ are concurrent, and similarly, we would have that $\{AA_1,CC_1,\ell\}; \{AA_1,BB_1,\ell\}$ are concurrent, implying the desired result. Now, we use Moving Points. Fix $A_1,B_1,C_1$ and move $P$ projectively on the incircle. Let $T_A=C'C_1 \cap B'B_1$. $\implies I$ is the incenter of $\Delta C_1T_AB_1$ $\implies \angle ZIY=90º+\frac{\angle FT_AE}{2}$ (fixed). Hence, the composed map $$C'C_1 \mapsto \mathcal{C}_I \mapsto \mathcal{C}_I \mapsto B'B_1$$given by $$C_1 \mapsto IC_1 \mapsto IB_1 \mapsto B_1$$is projective $(*)$, since it's a projection followed by a rotation with fixed angle through $I$, followed by another projection. Now, define $CC_1 \cap \ell= Q_C, BB_1 \cap \ell= Q_B$. Thus, the composed map $$C_1C' \mapsto \mathcal{C}_C \mapsto \mathcal{C}_C \mapsto \ell$$given by $$ A_1 \mapsto CA_1 \mapsto CQ_C \mapsto Q_C$$is projective, since all maps are projections, which are projective maps. Similarly, $Y \mapsto Q_B$ is projective $\implies$ from $(*)$, the map $$Q_B \mapsto B_1 \mapsto C_1 \mapsto Q_C$$is projective. Hence, since $Q_B,Q_C \in \ell$, which is a fixed line, by the Moving Points Lemma, in order to prove that $Q_B=Q_C$, it is sufficient to verify the problem for $3$ choices of $P$. Choosing $P$ to be the points such that the incircle touches $AB,BC,CA$, we get the desired result easily. Hence, we are done. $\blacksquare$

We will henceforth work in the projective plane. First of all send $\ell'$ to the line at infinity, so that $A', B', C'$ are the points at infinity for $BC, CA, AB$ respectively. Under this transformation, however the incircle becomes an ellipse. In order to fix this; we take an affine homography converting the ellipse into a circle again. Since affine transforms preserve the line at infinity, $\ell'$ remains the line at infinity. Now considering the projective dual of the problem, it can be seen to be equivalent to the following: Reduced Problem: Given a triangle $ABC$ with circumcircle $\gamma$, an arbitrary point $P$ on $\gamma$ and $A', B', C'$ as the respective $A, B, C$ antipodes with respect to $\gamma$ prove that $A'P \cap BC, B'P \cap CA$ and $C'P \cap AB$ are collinear. Now this is easy to prove by spamming Pascal's Hexagram Theorem. Using Pascal on $APCA'BC'$ yields that centre of $\gamma$, $A'P \cap BC$, $C'P \cap AB$ are collinear. Reiterating this result for other pairs of intersections yields that $A'P \cap BC, B'P \cap CA$ and $C'P \cap AB$ are collinear and thus we are done.

Polar reciprocate the entire diagram to get the equivalent problem statement. New Problem Statement wrote: Let $ABC$ be a triangle with point $X$ in the plane and point $P$ on circumcircle $\Gamma$. Let $XA, XB, XC$ intersect $\Gamma$ at $D, E, F$. Then let $DY, EY, FY$ intersect $BC, AC, AB$ at $G, H, I$. Show that $G, H, I$ are collinear. Now, take a homography in $\mathbb{CP}^2$ that maps $X$ to a point of infinity. Let $L$ be one of the midpoint of arcs $AD, BE, CF$. Oh wait, this is just USAMO 2012/5 with $\gamma = PL$.

brianchon on $ABDXYE,ACDXZF$ gives $AX,BY,CZ,DEF$ concurrent done

We apply Brianchon on $A_1A'BAB'B_1$ so we obtained that $AA_1,BB_1,l$ are concurrent. Analogous $BB_1,CC_1,l$ are concurrent so $AA_1,BB_1,CC_1$ are concurrent. So the problem is proved.

Denote by $\omega$ the incircle of $\triangle ABC$. We claim that in particular $ AA_1,BB_1,CC_1$ are concurrent on line $\ell$. Now, note that each side of $AC'C_1A_1A'C$ and $BC'C_1B_1B'C$ are tangent to $\omega$. Thus both these hexagons are tangential, which implies that by Brianchon's Theorem, $AA_1$ and $CC_1$ intersect on $\ell$ and also $BB_1$ and $CC_1$ intersect on $\ell$, which finishes the problem.