It follows from this configuration, which I believe I have seen before on this site (maybe hmm...) $\triangle ABC,$ point $D$ on segment $BC.$ $AD \cap (ABC) = T.$ Let $M, N$ be midpoint of $TB, TC$ resp. $AM, AN \cap (ABC) = E, F.$ $BF \cap CE = I, AE \cap BT=K, AF \cap CT = L,$ then $\overline{IKL} \parallel BC$ and $TA, TI$ are isogonal wrt $\angle BTC$ Main idea: call $J$ the midpoint of $TD,$ prove that $I, J$ are isogonal conjugate wrt $\triangle TBC$

$K=BP\cap AA$ and $L=CN\cap CC$. Notice that $KM=MB$ and $BN=NL$. By Pascal' s theorem applied to $CCAPBQ$, we have that $PM\cap CL$, $N=LQ\cap MA$ and $AP\cap QC$ are collinear, so $\triangle LQC$ and $\triangle MAP$ are in perspective, therefore $L-R-M$ collinear. Same way we can get $N-R-K$ collinear, so $BR$ must bisect $KL$, since $M$ and $N$ are midpoints of $KB$ and $LB$, implying that $BR$ also passes through the midpoint of $MN$.

Synthetic and elementary solution Let $X,Y$ be midpoints of $BA, BC$ and $G$ be the centroid of $\triangle ABC$. Notice that $\triangle BAH\cup M\sim\triangle CAB\cup X$ thus $\angle GCA = \angle MBA = \angle PCA = \angle RCA$. Analogously, we get $\angle GAC = \angle RAC$ therefore $R,G$ are symmetric across $AC$. Let $K$ be the midpoint of $AC$. And let $RG$ and $BR$ cut $AC$ at $U,V$ respectively, then since $RU = GU = \tfrac{1}{3}BH$, we get $UV : VH = 1:3$. However, $KU : UH = 1 : 2$ hence $V$ is the midpoint of $KH$ which implies that $V$ is also the midpoint of $MN$.

Let $B'$ be the reflection of $B$ across $AC$ and $D$ the antipode of $B'$ wrt $(ABC)$. Notice that $|AH||HN|=\frac{|AH||HC|}{2}=|MH||HC|$ and thus $H$ is the center of an involution on the line $AC$ swapping $(A,N)$ and $(C,M)$. Projecting from $B$ onto the circumcircle we obtain that $(A,Q)$, $(C,P)$ and $(B',D)$ are pairs under an involution on the circle, which again implies that $D$, $B'$ and $R=AQ\cap CP$ are collinear. Now note that $-1=(H,C;N,\infty_{AC})\overset{B}{=}(B',C;Q,D)\overset{A}{=}(B',O;R,D)\overset{B}{=}(H,O;BR\cap AC,\infty_{AC})$. Thus $BR\cap AC$ is the midpoint of $HO$. Here $O$ denotes the center of $(ABC)$. Finally note that $|MH|=\frac{|AH|}{2}=\frac{|AC|-|HC|}{2}=|OC|-|NC|=|ON|$, which implies that $MN$ and $HO$ indeed has a common midpoint.

I understand why $(A,N)$ and $(C,M)$ are involutions, but why is $(B',D)$ an involution in lebarelue's solution?

Plops wrote: I understand why $(A,N)$ and $(C,M)$ are involutions, but why is $(B',D)$ an involution in lebarelue's solution? Well, since $H$ is the center of the involution on $AC$, we have that $H$ is swapped with $\infty_{AC}$ under this involution. Thus, by projecting onto the circle, from $B$, we have that $H$ is sent to $B’$ while $\infty_{AC}$ is projected to $D$.

MarkBcc168 Can you explain why UV:VH=1:3? I've been trying to figure it out but I have no idea.

I think you are missing $BH\parallel UR$.

Thank you very much! Now I see

My solution with cartesian coordinates Let $A=(0,a),B=(0,0),C=(1,0)$ then we can easily get $H=(\frac{a^2}{a^2+1},\frac{a}{a^2+1})$ and so $M=(\frac{a^2}{2a^2+2},\frac{a^3+2a}{2a^2+2}),N=(\frac{2a^2+1}{2a^2+2},\frac{a}{2a^2+2})$ the equation of $(ABC)$ is : $x^2-x+y^2-ay=0$ and the equation of the two lines : BM: $y=\frac{a^2+2}{a}x$ , BN: $y=\frac{a}{2a^2+1}x$ and by solving the system of equations we get: $P=(\frac{a^4+3a^2}{a^4+5a^2+4},\frac{a^5+5a^3+6a}{a^4+5a^2+4})$ and $Q=(\frac{6a^4+5a^2+1}{4a^4+5a^2+1},\frac{3a^3+a}{4a^4+5a^2+1})$ and the equation of the two lines : AQ: $y=\frac{-4a^5-2a^3}{6a^4+5a^2+1}x+a$ CP: $y=\frac{a^5+5a^3+6a}{2a^2+4}(1-x)$ and also by solving the system of equations: $R=(\frac{3a^2+1}{3a^2+3},\frac{a^3+3a}{3a^2+3})$ and then the equation of the line $BR$ is: BR: $y=\frac{a^3+3a}{3a^2+1}x$ now if we let $D$ to be the midpoint of $MN$ : $\implies D=(\frac{3a^2+1}{4a^2+4},\frac{a^3+3a}{4a^2+4})$ and clearly $D$ satisfy the equation of the line $BR$ and so we are done.

A nice problem Let$ E=AA\cap BM , F=CC\cap BN $ $\color{red} \textbf{claim:}$ $E,R,N$ are collinear $\color{green} \textbf{proof:}$ Pascal's theorem on $AQBPCA$ $\blacksquare$ $\color{red} \textbf{claim:}$ $M$ is the midpoint of $BE$ $\color{green} \textbf{proof:}$ $\triangle MAE\sim \triangle MHB ,AM=HM$ $\blacksquare$ Similar arguments for $F$ Now $R$ is the centroid of $\triangle BEF$ , by a dilation at $B$ we are done.

Took about 45 minutes. Quite nice, I have to say. Let S be the intersection of the line parallel to AC passing through B and PC. T is the intersection of BQ and PC. X is the intersection of AC and BR. We need to prove that (S, R; T, P) = -1, because when we have (S, R; T, P) = (infinity, X; N, M), MX=N and we're done. So, everything what we need is to show PR/RT = PS/ST. Notice that PS = PC*(BS/SM), ST = CT*(BS/CN). So we have PS/ST = PC/CT * CN/CM Then we have PR = RQ*(AR/RC) and by Menelaus in triangle ACR we also have 1/RT = AQ/RQ * CN/AN * 1/CT. So, we received that PR/RT = AR/CR * CN/AN * AQ/CT. We need to show that AR/CR * CN/AN * AQ/CT = PC/CT * CN/CM, which is equal to AR/CR * AQ/AN * CM/CP =1. This is trivial by law of sines.