any solution

Hint: we can use feuerbach's theorem:the nine point circle is tangent to incircle.

Yep, but I think you should correct the statement as it is false.

Let me restate problem first to nicer formulation. Problem: Let $\triangle ABC$ be given and it's incircle $(I)$. Let $(I)$ touches sides $AB$, $BC$ and $AC$ at $F$, $D$ and $E$, respectively and denote by $P$, $M$ and $N$ midpoints of same sides, respectively. Let $AM\cap PN=\{ J\}$ and $AD\cap BJ=\{ K\}$. If circle centered at $C$ with radius $AC$ cuts $BC$ at $X$, line parallel to $BC$ through $K$ meets $AX$ at $U$ and $FU\cap BC=\{ R\}$, then prove that $Q$ lies on tangent from Feuerbach point $F_e$ onto $(I)$. Notation: Let $A_H$ be foot of altitude from $A$ on $BC$, let $R$ be intersection point of tangent from $F_e$ on $(I)$ with $BC$ and denote with $a$, $b$ and $c$ sides $BC$, $AC$ and $AB$. Proof: Assume WLOG that $b>a$ Let $R'$ be harmonic conjugate of $R$ WRT $A_HM$ i.e. $$-1=(R,\ R';\ A_H,\ M)\Longleftrightarrow \frac{RA_H}{RM}=\frac{R'A_H}{R'M}=\frac{F_eA_H^2}{F_eM^2}$$But from $\text{Casey's}$ we have \begin{align*} \left(\frac{F_eA_H}{F_eM}=\frac{A_HD}{MD}\right)^2& \Longrightarrow \frac{RA_H}{RM}=\frac{A_HD^2}{MD^2}\\ & \Longleftrightarrow \frac{MA_H}{RM}=\frac{MD^2-A_HD^2}{MD^2}\\ & \Longleftrightarrow RM=\frac{MD^2\cdot MA_H}{(MD-A_HD)\cdot MA_H}\\ & \qquad \qquad \ \ = \frac{MD^2}{MD-A_HD}\\ & \qquad \qquad \ \ = \frac{MD^2}{2MD-A_HM}\\ & \qquad \qquad \ \ = \frac{\left(\frac{a-(a+c-b)}{2}\right)^2}{(a-(a+c-b))-A_HM}\\ & \qquad \qquad \ \ = \frac{\frac{(b-c)^2}{4}}{b-c-\frac{b^2-c^2}{2a}}\\ & \qquad \qquad \ \ = \frac{a(b-c)}{2(2a-b-c)} \end{align*} Now let's finish it! \begin{align*} \begin{rcases} RB=RM- BM=\frac{a(b-a)}{2a-b-c}\\ RX=RB-(b-a)=\frac{(a-b)(a-b-c)}{2a-b-c}\\ BF=\frac{a+c-b}{2}\\ AF=\frac{b+c-a}{2}\\ \frac{AU}{UX}\stackrel{UK||XD}{=}\frac{AK}{KD}\stackrel{\text{Menelaus on }\triangle ADM}{===}\frac{BM}{BD}\cdot \frac{AJ}{JM}\stackrel{AJ=JM}{=}\frac{BM}{BD} =\frac{a}{a+c-b} \end{rcases}\Longrightarrow \frac{RX}{RB}\cdot \frac{BF}{FA}\cdot \frac{AU}{UX}&=\frac{\frac{(a-b)(a-b-c)}{2a-b-c}}{\frac{a(b-a)}{2a-b-c}}\cdot \frac{\frac{a+c-b}{2}}{\frac{b+c-a}{2}}\cdot \frac{a}{a+c-b}\\ & =1 \end{align*}And by Menelaus we are done!

Sorry for revive, but is there any proof without calculations?

KereMath wrote: Hint: we can use feuerbach's theorem:the nine point circle is tangent to incircle. Hortlattım sorry Post solution without calc reco

İnmemories you want too much things bro. It`s Fehmi Kadan`s question. What dou you expect?

Kyolcu22 wrote: İnmemories you want too much things bro. It`s Fehmi Kadan`s question. What dou you expect? Something synthetic

Once we see that all points are actually well-described points, what can stop us from bary bashing?? First, let's see that our point $P$ must satisfy $PG^2=PD\cdot PH$, where $H$ is the foot of altitude from $A$. Also, a point satisfying this condition is unique. Remember that some of the coordinates in our solution are non-normalized: $D=(0:1/2:1/2)$, $J=(1:1/2:1/2)$ and $G=\left( 0: \frac{s-c}{a} : \frac{s-b}{a}\right)$ so writing the equation of $$AG: \frac{s-c}{a}\cdot z - \frac{s-b}{a}\cdot y = 0 \text{ and } BJ : x-2z=0$$We deduce that $K=(2\cdot(s-b) : s-c: s-b)$ and the equation of the line parallel to $BC$ is $ax -2(s-b)(y+z)=0$ . Note that $X=(0:b:a-b)$ and the equation of line $AX: bz -(a-b)\cdot y=0$ From these, we get $U=(2(s-b): b: a-b)$ and we know that $I= \left( \frac{s-b}{c} : \frac{s-a}{c} : 0\right)$, so $$UI : -(s-a)(a-b)x + (s-b)(a-b)y + (s-b)(c-a)z = 0$$deducing $P=(0 : a-c : a-b)$, so its normalized coordinates are $P= \left(0 , \frac{a-c}{2a-b-c}, \frac{a-b}{2a-b-c}\right)$. Also note that normalized coordinates of $H = \left(0 , \frac{a^2+b^2-c^2}{2a^2}, \frac{a^2-b^2+c^2}{2a^2}\right)$ With these coordinates, we can find the vectors $\overrightarrow{PG}$, $\overrightarrow{PD}$ and $\overrightarrow{PH}$. All these vectors lie on $BC$ so the lenghts are basically $-a^2yz$ for any $\overrightarrow{PQ} = (x,y,z)$, $Q \in \{G,D,H\}$, because of the fact that $x=0$. The rest is just some annoying computations. If I didn't make a typo, desired equality holds for these coordinates

BUMPing for a synthetic solution

inmemories wrote: Kyolcu22 wrote: It`s Fehmi Kadan`s question. What dou you expect? Something synthetic I think Fehmi Kadan does not always propose extremely difficult or bashy problems, but of course sometimes he did ask harder ones. In fact, geometry solutions in the booklets of Turkish National Olympiad (the official source) never used advanced/ultimate concepts, for example, barycentric coordinates or complex numbers or just plain trigonometry.