This problem can be easily solved by using complex numbers Actually this problem is same as Russian M.O. 1998. Elementary solution can be found in in the book "Problems and solutions from national olympiads arround the world 1998-1999" published by MAA (Titu Andreescu and Zuming Feng). Namely it is Russia 1998 problem 39 (page 130)

I don't have that book and this problem was bugging me for some time now, and now I am happy that I found a nice synthetic-projective solution: First, $\angle{ABK_{b}}=\frac{\angle{B}}{2}+\angle{EBK_{b}}$. Now $\angle{K_{a}K_{b}B}=\angle{LK_{b}B}-\angle{LK_{b}K_{a}}$ $(\star)$, where $L \in EE \cap DD$. But $\angle{LK_{b}B}=180-\angle{EK_{b}B}=$ $=180-(180-\angle{EBK_{b}}-\angle{BEK_{b}}) = \angle{EBK_{b}}+\angle{BEK_{b}}$. And $\angle{BEK_{b}}=\angle{BEA}=180-\angle{A}-\frac{\angle{B}}{2}=\angle{C}+\frac{\angle{B}}{2}$ Now on other hand, $\angle{LK_{b}K_{a}}=\frac{180-\angle{ELD}}{2}=$ $=\frac{180-(540-\angle{A}-\angle{B}-2\angle{BEA}-2\angle{BDA})}{2}=\angle{C}$. Thus, by replacing it in $(\star)$ we have that $\angle{K_{a}K_{b}B}=\angle{ABK_{b}}$, so $AB \| K_{a}K_{b}$, therefore $GH \| K_{a}K_{b}$ (where $(HG)$ is the midline coresponding to $(AB)$, i.e $G,H,J$ are the midpoints of the sides) Now, in a similar way, $JH \| K_{b}K_{c}$, and $GJ \| K_{c}K_{a}$. Therefore, the points $GH \cap K_{a}K_{b}=\infty$, $JH \cap K_{b}K_{c}=\infty$, $GJ \cap K_{c}K_{a}=\infty$, are "collinear", so by Desargues' theorem we have that the lines $GK_{a}, HK_{b}, JK_{c}$(the desired lines) are concurrent in a point $M$. (I guess it had been resulted more trivially since observing the paralellism, without using Desargues) Now for proving that $M$ in on the incircle, I got nastily complicated, but I discovered a more general result. Denote $GG_{a}$, the tangent from the midpoint $G$ to the incircle ($K \neq A'$), where $A'$ is the point of tangenty of the incircle with $BC$. Now the cyclic quadrilateral $G_{a}K_{a}A'M*$ is harmonical, where $M*$ is the intersection of $GK_{a}$ with the incircle. Therefore $A(G_{a}K_{a}A'M*)$ is an harmonical pencil, and now construct the similar ones $B(G_{b}K_{b}B'M**)$ and $C(G_{c}K_{c}C'M***)$. It is clear that if 3 series of lines are concurrent, the fourth is concurrent. Therefore, if the only thing remaining is to prove that the series of lines $(AA', BB',CC')$ , $(AK_{a}, BK_{b}, CK_{c})$, $(AG_{a}, BG_{b}, CG_{c})$ are concurent. The first two are trivially true, because $(AA', BB',CC')$ are concurrent in the Gergonne point, and the $AK_{a}$ type lines are the isogonals of the $AA'$ type lines, so they are also concurrent. Now for the concurrency of the $AG_{a}$ type lines I have the following generalization:("last lemma" drawing) *** Let $ABC$ be a triangle and $P$ a point inside it, and consider the intersections $A_{1}, B_{1}, C_{1}$, of the lines $AP, BP, CP$ with the opposite sides. Denote by $A_{1}X$ the second tangent from $A_{1}$ w.r.t to the incircle and the other similar ones. Prove that the $AX$ type lines are concurrent. For this superb result, the proof I had in mind was, by noticing that it is suffice to consider $P$ inside the incircle, and for this, it is an projective transformation that maps the incircle to its self, and allows $P$ to be anywhere else in the incircle, i.e it's center. A proof for this projective trasnformation, I think that can be found in Prasolov. Thus, by considering $P \equiv I$, we have that the $AX$ type lines are mapped into the isogonals of the Gergonne-type lines $AA_{1}$, so they are concurrent. Therefore, by returning in our problem we have that for $P \equiv G$ (the centroid), the $AG_{a}$ type lines are concurrent, so also the lines $AM*, BM**, CM***$ are concurrent in $M$, (because we have considered that the $M* \in AM \cap w$ ($w$ = incircle) But because, the $GK_{a}$ type lines are also concurent in $M$, it trivially results that $M\equiv M* \equiv M** \equiv M***$, so the problem is solved. Remark: From the "last lemma", and from the Desargues-solution for the concurrenty of the $GK_{a}$ type lines, we can see that the problem can be a little generalized, such that $G,H,J$ are not anymore the midpoints of the sides of the triangle, but they respect the Desargues colliniarity in the beggining of the solution.

Attachments:

See the lemma from http://www.mathlinks.ro/Forum/viewtopic.php?t=153668

Maybe this link can be helpful for you. http://www.dms.org.yu/prilozi/Zadaci_resenja_SMO.pdf Second solution in 5. problem is with complex numbers and it was assumed that incircle of triangle $ABC$ is unit circle in complex field.

Can you please give a link of this solution, that is not writen is cirilica, so that i can read it, because i cant open the given link.

An equivalent statement of the problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195473

Hint: $ A_{1}K_{a},B_{1}K_{b},C_{1}K_{c} $ intersect on the Feurbauch point of $ ABC $.

Dear Mathlinkers, this problem coming from Fontené has been proposed in 1982 (13th OIM) Sincerely Jean-Louis

Claim I: $\triangle A_1B_1C_1$ and $\triangle K_AK_BK_C$ are similar and homothetic Claim II: $A_1K_A,B_1K_B,C_1K_C $ concur at the Feurbauch point of $\triangle ABC$ this is because the $N_9$ circle passes through the points $A_1,B_1$ and $C_1$. Therefore the center of homothety should be the Feuerbach point ($X_{11}$). The rest is just angle chasing for which choose an arbitary point on the incircle and continue....