I am surprised because of it was a contest problem ! Any way, we can get a solution of the wanted result, applying the proof of the generalization of a familiar problem in the topic Interesting and hard, of cvix. Kostas Vittas.

Dear Kostas, Why are you surprised?

Can someone write out a solution?I can't continue vittasko's comment...

Omid Hatami wrote: Let $ A,B,C,Q$ be fixed points on plane. $ M,N,P$ are intersection points of $ AQ,BQ,CQ$ with $ BC,CA,AB$. $ D',E',F'$ are tangency points of incircle of $ ABC$ with $ BC,CA,AB$. Tangents drawn from $ M,N,P$ (not triangle sides) to incircle of $ ABC$ make triangle $ DEF$. Prove that $ DD',EE',FF'$ intersect at $ Q$. we just need prove $CP,NB$ and $D'D$ are concurrent at $Q$ Tangents drawn from $ N,P$ (not triangle sides) to incircle of $ ABC$ intersection $AB, AC at J , K$ and intersection incircle of $ ABC$ at $O,V$ *) $PKCB$ is a special quadrilateral ( i don't know this name ) and we know following lemma: $F'O$ intersection $D'E'$ at $X$ then $X$ is pole of $PC.$ *) $NJBC$ is a special quadrilateral and $E'V$ intersection $D'F'$ at $Y$ then $Y$ is pole of $BN$ *) $OV$ intersection Tangents drawn from $D'$ to incircle of $ ABC$ at $Z$ then $Z$ is pole of $DD'$ and use Pascal theorem for $(D'D'F'VOE')$ we have $XYZ$ therefore $DD',PC,BN$ are concurrent Image not found

The idea of using pole-polar and Pascal is good, but additionally I think the whole problem is rested on nothing else than two well known properties of pole-polar: La Hire’s Theorem (Let x and y be the polars of X and Y, respectively; then X is on line y ⇔ Y is on line x) and this: Let x, y, z be the polars of distinct points X, Y, Z, respectively; then Z = x∩y ⇔ z = XY (what can be proved easily using La Hire). Let $E''F''$ meet $BC$ at $G$, $E'E''$ meet $F'D'$ at $H$, $F'F''$ meet $D'E'$ at $K$. From now on, while talking about poles and polars we do so with respect to the incircle of $ABC$. First notice that $E''F''$ is the polar of $D$, as $G$ lies on the polar of $D$, $D$ must lie on the polar of $G$, but we know that the tangent point $D'$ is also on the polar of $G$, so $DD'$ is the polar of $G$. Then to show that $DD'$ passes through $Q$, it suffices to show that $Q$ lies on the polar of $G$. Now notice that $F'F''$ is the polar of $P$ and $D'E'$ is the polar of $C$, then $PC$ is the polar of $K\equiv F'F''\cap D'E'$. $E'E''$ is the polar of $N$ and $F'D'$ is the polar of $B$, then $NB$ is the polar of $H$. Then $HK$ is the polar of $Q\equiv PN\cap BN$. Using Pascal for $E''F''F'D'D'E'$ we get $G\equiv E''F''\cap D'D', K\equiv F''F'\cap D'E', H\equiv E''E'\cap F'D'$ collinear. As $G$ lies on the polar of $Q$, $Q$ must lie on the polar of $G$. q.e.d. By the same way we get $EE'$ and $FF'$ pass through $Q$. Note: Having a look back at the cvix's interesting and hard vittasko mentioned above, we see that the problem can be solved in the same way, of course with the aid of a well known lemma (instead of Pascal in our problem).

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We can use brianchon's theorem on $BD'CNDP$ then we get Q, D, D' are collinear, it finishes the proof.