Let $H_A, H_B, H_C$ be the feet of the altitudes from $A, B, C$ in $\triangle{ABC}$, and let $H_B', H_C'$ be the intersections of $BH, CH$ with the line through $A$ parallel to $BC$; then $(P, H_C; H, H_C') = (Q, H_B; H, H_B') = -1 \implies$ by the Prism Lemma, $H_BH_C, PQ, H_B'H_C'$ concur at point $X$. Projecting $(P, H_C; H, H_C')$ from $X$ onto line $AH$, we get a harmonic bundle, and by Ceva-Menelaus $(A, H; H_BH_C \cap AH, H_A) = -1 \implies XP \cap AH = H_A$, and so the three lines concur at $H_A$. $\square$

v4913 wrote: Let $H_A, H_B, H_C$ be the feet of the altitudes from $A, B, C$ in $\triangle{ABC}$, and let $H_B', H_C'$ be the intersections of $BH, CH$ with the line through $A$ parallel to $BC$; then $(P, H_C; H, H_C') = (Q, H_B; H, H_B') = -1 \implies$ by the Prism Lemma, $H_BH_C, PQ, H_B'H_C'$ concur at point $X$. Projecting $(P, H_C; H, H_C')$ from $X$ onto line $AH$, we get a harmonic bundle, and by Ceva-Menelaus $(A, H; H_BH_C \cap AH, H_A) = -1 \implies XP \cap AH = H_A$, and so the three lines concur at $H_A$. $\square$ imagine. Use coordinates with $A=(0,1),B=(b,0),C=(c,0)$. Then $\overline{CH}$ has equation $y=bx-bc$ and $\ell_b$ has equation $y=\tfrac{1}{b}x+1$, so we can calculate $P=(\tfrac{b^2c+b}{b^2-1},\tfrac{b^2+bc}{b^2-1})$. The line through the origin (which is just $\overline{AH} \cap \overline{BC}$) and $P$ has slope $\tfrac{b+c}{bc+1}$. Since this is symmetric in $b$ and $c$, it follows that $Q$ lies on this line as well. $\blacksquare$

See CJMO 2021/1.

IAmTheHazard wrote: Use coordinates with $A=(0,1),B=(b,0),C=(c,0)$. Then $\overline{CH}$ has equation $y=bx-bc$ and $\ell_b$ has equation $y=\tfrac{1}{b}x+1$, so we can calculate $P=(\tfrac{b^2c+b}{b^2-1},\tfrac{b^2+bc}{b^2-1})$. The line through the origin (which is just $\overline{AH} \cap \overline{BC}$) and $P$ has slope $\tfrac{b+c}{bc+1}$. Since this is symmetric in $b$ and $c$, it follows that $Q$ lies on this line as well. $\blacksquare$ imagine DDIT from $A$ onto the complete quadrilateral formed by $\overline{CH},\overline{BH},\overline{BC},\overline{PQ}$ means that if $K=\overline{BC}\cap\overline{PQ}$, the pairs $(\overline{AC},\overline{AQ}),(\overline{AB},\overline{AP}),(\overline{AK},\overline{AH})$ form an involution, which by the first two pairs must be the reflection around $\overline{AH}$, so $\overline{AK}\equiv \overline{AH}$ as desired.

^ sniped o_o? [as i have already this proposal in mind for a long time, only using it now in a contest]

navi_09220114 wrote: ^ sniped o_o? [as i have already this proposal in mind for a long time, only using it now in a contest] To clarify, I'm not accusing you of plagiarism. It's totally possible that two different proposers came up with this precise idea, as it's quite natural in my opinion. The purpose of posting the other thread was to provide additional solutions and commentary.

Ah okay, no worries

Let $B'=AP \cap BC, C'=AQ \cap BC$ and $D = AH \cap BC$. Notice that because of symmetry we have that: $ \angle AC'B = \angle ACB = 180 - \angle AHB$, hence $C'$ lies on $(AHB)$ and by power of point we get that $QC' \cdot QA = QB \cdot QH$, so $Q$ lies on the radical axis of $(BHC)$ and $(AB'C')$. Analogoulsy $P$ lies on the radical axis of $(BHC)$ and $(AB'C')$. Finally because of symmetry we get that $DB=DB'$ and $DC=DC'$, so $DB' \cdot DC' = DB \cdot DC$, hence $D$ also lies on the radical axis of $(BHC)$ and $(AB'C')$, which is $PQ$.

Attachments:

It's easy to see that the lines $AB$ and $AP$ are isogonal with respect to triangle $AQC$, and that $AH$ is the bisector of angle $QAC$ due to the reflexion. Let $D$ be the intersection between $QP$ and $BC$. Now, by the isogonal line lemma, $QB\cap CP$ and $QP\cap CB$, which are the points $H$ and $D$ respectively, are also isogonal. As $H$ is in the bisector of $QAC$, so must be $D$, giving the $AHD$ collinearity

An alternate(bad?) solution using radical axes and lots of construction, probably this is what happens when a geometer does geometry when tired. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.26577412414945, xmax = 11.862177815648126, ymin = -8.254776995461313, ymax = 3.8363443170128306; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((3.32,-0.11)--(0.9800641531553963,-5.769451412508511)--(11.22,-5.73)--cycle, linewidth(0) + blue); /* draw figures */ draw((xmin, -259.55815510121175*xmin + 861.6230749360229)--(xmax, -259.55815510121175*xmax + 861.6230749360229), linewidth(0.4)); /* line */ draw((xmin, 2.4186352887153997*xmin-8.139869158535125)--(xmax, 2.4186352887153997*xmax-8.139869158535125), linewidth(0.4)); /* line */ draw((xmin, -0.7113924050632912*xmin + 2.2518227848101264)--(xmax, -0.7113924050632912*xmax + 2.2518227848101264), linewidth(0.4)); /* line */ draw((xmin, -2.3668197441980086*xmin + 7.747841550737389)--(xmax, -2.3668197441980086*xmax + 7.747841550737389), linewidth(0.4)); /* line */ draw((xmin, 0.7230614888547111*xmin-2.5105641429976404)--(xmax, 0.7230614888547111*xmax-2.5105641429976404), linewidth(0.4)); /* line */ draw((xmin, -0.41345630102466824*xmin-1.0910203025032221)--(xmax, -0.41345630102466824*xmax-1.0910203025032221), linewidth(0.4)); /* line */ draw((xmin, 1.4056939501779364*xmin-7.147121663385315)--(xmax, 1.4056939501779364*xmax-7.147121663385315), linewidth(0.4)); /* line */ draw((3.341769119397886,-5.760352469093222)--(6.792172630349427,2.400594311661035), linewidth(0.8) + linetype("4 4") + ccqqqq); draw(circle((8.444509199459985,-1.2689995238907779), 5.253938944423418), linewidth(0.4)); draw((xmin, 0.0038527011397895996*xmin-5.773227306788439)--(xmax, 0.0038527011397895996*xmax-5.773227306788439), linewidth(0.4)); /* line */ draw(circle((-1.7954266473846316,-1.308450936399283), 5.253938944423422), linewidth(0.4)); draw(circle((6.806334794626311,-1.2753109202874537), 3.675932513140479), linewidth(0.4) + qqwuqq); draw(circle((2.154573352615375,-4.118450936399285), 2.0261476826058384), linewidth(0.4) + ccqqqq); draw(circle((-1.809588811661518,2.3674542955492055), 8.601825281062972), linewidth(0.4) + ccqqqq); draw(circle((-0.608230880602121,-2.950352469093221), 4.8475354562911726), linewidth(0.4) + qqwuqq); draw((xmin, -1.3830082439931133*xmin + 2.136698154163788)--(xmax, -1.3830082439931133*xmax + 2.136698154163788), linewidth(0.4)); /* line */ draw(circle((3.324541276037679,-1.2887252301450307), 1.1787339782022517), linewidth(0.4) + ccqqqq); draw((-4.53646176120424,-5.790704938186443)--(4.166543764511911,-2.1136164961745036), linewidth(0.4)); /* dots and labels */ dot((3.32,-0.11),dotstyle); label("$A$", (3.388529221833601,0.06223708652957186), NE * labelscalefactor); dot((0.9800641531553963,-5.769451412508511),dotstyle); label("$B$", (1.0471849214412037,-5.598923759195316), NE * labelscalefactor); dot((11.22,-5.73),dotstyle); label("$C$", (11.28619805599303,-5.563978321876027), NE * labelscalefactor); dot((3.3290825520753557,-2.4674504602900615),linewidth(4pt) + dotstyle); label("$H$", (3.4060019404932453,-2.331525369841754), NE * labelscalefactor); dot((4.524944850448016,-2.9618872627100794),linewidth(4pt) + dotstyle); label("$P$", (4.5941468093490885,-2.820761492311806), NE * labelscalefactor); dot((6.792172630349427,2.400594311661035),linewidth(4pt) + dotstyle); label("$Q$", (6.865600235102907,2.5433631361991216), NE * labelscalefactor); dot((3.341769119397886,-5.760352469093222),linewidth(4pt) + dotstyle); label("$D$", (3.4060019404932453,-5.616396477854961), NE * labelscalefactor); dot((5.703474085640374,-5.7512535256779325),linewidth(4pt) + dotstyle); label("$E$", (5.764818959545288,-5.616396477854961), NE * labelscalefactor); dot((-4.53646176120424,-5.790704938186443),linewidth(4pt) + dotstyle); label("$F$", (-4.474194175006539,-5.6513419151742506), NE * labelscalefactor); dot((4.159529315080045,-3.8261736360381726),linewidth(4pt) + dotstyle); label("$M$", (4.2446924361561935,-4.026379079827292), NE * labelscalefactor); dot((2.4889198080909716,-2.1200798799035416),linewidth(4pt) + dotstyle); label("$I$", (2.567311444830297,-1.9820709966488599), NE * labelscalefactor); dot((4.166543764511911,-2.1136164961745036),linewidth(4pt) + dotstyle); label("$J$", (4.2446924361561935,-1.9820709966488599), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $E,F$ denote the reflections of $B,C$ in $D$ respectively. Let $M$ be the miquel point of $AHBF$, $J$ be on $AE$ s.t. $FHJ \perp AE$. Note that It suffices to prove $P-M-D$, but this follows immediately from radax on $(AIHJ),(BHDIM),(AJMDF)$ (last two are cyclic from miquel).