K? You aren't note where is K or write K mistakenly

Is Vietnam TST 2020? number problem?

I think the question should be In acute $\triangle ABC$, $O$ is the circumcenter, $I$ is the incenter. The incircle touches $BC,CA,AB$ at $D,E,F$. And the points $K,M,N$ are the midpoints of $BC,CA,AB$ respectively. a) Prove that the lines passing through $D,E,F$ in parallel with $IK,IM,IN$ respectively are concurrent. b) Points $T,P,Q$ are the middle points of the major arc $BC,CA,AB$ on $\odot O$. Prove that the lines passing through $D,E,F$ in parallel with $IT,IP,IQ$ respectively are concurrent. I think các cung lớn is the major arc. So $T,P,Q$ are not the intersections of $AI,BI,CI$ and $\odot O$.

@YII.I. Yes I did modified the question so it sounds more natural for everyone. Straight translation is not so okay since we have many terminologies and assumptions that is not well known everywhere else. YII.I. wrote: I think các cung lớn is the major arc. So $T,P,Q$ are not the intersections of $AI,BI,CI$ and $\odot O$. @Above: I just read the question again and found that you are right about that. I will adopt your translation as problem statement.

gausskarl wrote: In acute $\triangle ABC$, $O$ is the circumcenter, $I$ is the incenter. The incircle touches $BC,CA,AB$ at $D,E,F$. And the points $K,M,N$ are the midpoints of $BC,CA,AB$ respectively. a) Prove that the lines passing through $D,E,F$ in parallel with $IK,IM,IN$ respectively are concurrent. b) Points $T,P,Q$ are the middle points of the major arc $BC,CA,AB$ on $\odot ABC$. Prove that the lines passing through $D,E,F$ in parallel with $IT,IP,IQ$ respectively are concurrent. Solution for a). Let $D'$ be the tangency point of the $A$-excircle and $BC$. Define $E',\ F'$ analogously for the $B$ and $C$-excircles, respectively. Recall that $AD',\ BE'$ and $CF'$ concur at the Nagel point of $\bigtriangleup ABC$, say $J$, and that $K$ is the midpoint of $\overline{DD'}$. Moreover, $DI$ and $AD'$ meet each other at the antipode of $D$ on the incircle, say $D''$. Clearly, $KI\parallel D'D''$. Let $J'$ be the symmetric point of $J$ with respect to $I$. Since $J'DJD''$ is a parallelogram, we have $DJ'\parallel D''D'\parallel IK$. Similarly, we can show that $J'E\parallel IM$ and $J'F\parallel IN$, thus $J'$ is the required concurrency point. $\square$ Solution for b). As usual, define $T_A$ to be the $A$-mixtilinear intouch point. $T_B$ and $T_C$ are defined similarly. It's known that $AT_A,\ BT_B$ and $CT_C$ concur at the exsimilicenter $T''$ of the incircle and the circumcircle of $\bigtriangleup ABC$. Let $T',\ P'$ and $Q'$ be the intersection points of rays $T_AA,\ T_BB$ and $T_CC$ with the incircle, respectively. Suppose that $D'',\ E''$ and $F''$ are the antipodes of $D,\ E$ and $F$ on the incircle, respectively. It's not hard to prove that $AD''IT'$ is cyclic (take into account that $T_A$ is the center of spiral similarity carrying $\overline{AI}$ to $\overline{ID}$ and do some angle-chasing). Analogously, quadrilaterals $BE''IP'$ and $CF''IQ'$ are cyclic. By definition of $T''$, we know that $T_BT_C\parallel P'Q'$, so $BQ'P'C$ is cyclic by Reim's theorem. Similarly, $AQ'T'C$ and $AP'T'B$ are cyclic as well. Hence, $AT''\cdot T''T'=BT''\cdot T''P'= CT''\cdot T''Q'$. The latter implies that $T''$ has equal powers with respect to $(AD''IT'),\ (BE''IP')$ and $(CF''IQ')$, which leads us to conclude that these circles are coaxal; define $X$ as their other common point. We also infer that $X,\ T,\ I$ and $O$ are collinear. Furthermore, by applying the radical axis theorem to these three circles and the incircle, three at once, we conclude that $T'D'',\ P'E''$ and $Q'F''$ concur at a point on $OX$, say $Z$. Finally, recall that $IT_A$ is the perpendicular bisector of $\overline{T'D}$ ($T'$ and $D$ are symmetric with respect to the internal bisector of $\angle BT_AC$, the line $T_AI$, which is known to pass through $T$). Because $\angle D''T'D=90^\circ$, we get $D''T'\parallel TI$. These facts imply that $DZ'\parallel IT$, where $Z'$ is the symmetric point of $Z$ with respect to $I$. The same conclusion holds for $EZ',\ IP$ and $FZ',\ IQ$, so we are done $\square$ .

who has the rest of the problems please?

For part a) use homothety twice and then just consider reflection of nagel point over $I$ to win. For part b), first let $H_D$ a point in the incircle such that $DH_D \perp EF$, then let $Q_D$ be the D-queue point of $\triangle DEF$, and let $D'$ the opposite diametral of $D$ on the incircle, let $S_A$ the A-sharkydevilpoint and let $AS_A \cap BC=X_A$. First notice that $IT \parallel Q_DD'$, cause by trivial orthocenter config we get $Q_DEH_DF$ is an harmonic quadrilateral so $A,H_D,Q_D$ are colinear, and then since on spiral sim from $S_A$ we quickly check that $H_D \to I$ which means that $H_DQ_D \to IT$ (the lines) so all we need to verify is that $\angle H_DQ_DD'=\angle IS_AH_D$ from the spiral sim, but this is easy, as one can easly check that $S_A, DH_D \cap EF$ are inverses on the incircle thus $S_AH_DID$ are cyclic, but from radax we can also get $AI \perp IX_A$ so $IX_A \perp DH_D$ which means that $X_AS_AH_DID$ is cyclic and arcs $H_DI, ID$ are equal on that circle meaning that $\angle IS_AH_D=\angle IX_AD$ which after spiral sim is on the incircle half of arc $EH_D$ minus arc $FH_D$ but as diameter, altitude are isogonal this is just half of arc $H_DD'$ which is what we wanted. So to finish just recall from queue config that $Q_DD'$ goes through $H$ the orthocenter of $\triangle DEF$ so the point of concurrency desired on the problem is simply reflection of $H$ over $I$, thus we are done .