[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -4.611923266584953, xmax = 32.08728017846332, ymin = -14.68999279614231, ymax = 9.545779868542146; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((7.06,1.92)--(3.28,-5.96), linewidth(1.2) + rvwvcq); draw((7.06,1.92)--(14.88,-9.52), linewidth(1.2) + rvwvcq); draw((14.88,-9.52)--(3.28,-5.96), linewidth(1.2) + rvwvcq); draw((1.719054099368246,1.5123943424432678)--(7.766458399671943,-7.336878612313112), linewidth(1.2) + dbwrru); draw((12.400945900631754,2.327605657556732)--(7.766458399671943,-7.336878612313112), linewidth(1.2) + dbwrru); draw((1.719054099368246,1.5123943424432678)--(14.88,-9.52), linewidth(1.2) + dbwrru); draw((3.28,-5.96)--(12.400945900631754,2.327605657556732), linewidth(1.2) + dbwrru); draw((4.742309275832898,-2.9115880704859154)--(10.084044692813741,-2.503922159308081), linewidth(2) + dbwrru); draw((7.06,1.92)--(6.80847719440843,-2.7539040668014776), linewidth(1.2) + green); draw((7.06,1.92)--(9.08,-7.74), linewidth(1.2) + green); draw((1.719054099368246,1.5123943424432678)--(12.400945900631754,2.327605657556732), linewidth(1.2) + dtsfsf); draw((1.719054099368246,1.5123943424432678)--(3.28,-5.96), linewidth(1.2) + dtsfsf); draw((12.400945900631754,2.327605657556732)--(14.88,-9.52), linewidth(1.2) + dtsfsf); draw((7.06,1.92)--(7.766458399671943,-7.336878612313112), linewidth(1.2) + green); /* dots and labels */ dot((7.06,1.92),dotstyle); label("$A$", (7.1604186177357265,2.1582778763571104), NE * labelscalefactor); dot((3.28,-5.96),dotstyle); label("$B$", (3.3713450152924302,-5.7296677991694915), NE * labelscalefactor); dot((14.88,-9.52),dotstyle); label("$C$", (14.976872338499255,-9.280434885735847), NE * labelscalefactor); dot((1.719054099368246,1.5123943424432678),dotstyle); label("$P$", (1.822352662092341,1.7531567993663182), NE * labelscalefactor); dot((7.766458399671943,-7.336878612313112),linewidth(4pt) + dotstyle); label("$D$", (7.851507513778843,-7.135676242843418), NE * labelscalefactor); dot((12.400945900631754,2.327605657556732),dotstyle); label("$Q$", (12.498484573379113,2.5633989533479027), NE * labelscalefactor); dot((4.742309275832898,-2.9115880704859154),linewidth(4pt) + dotstyle); label("$F$", (4.848845413729439,-2.7270056991200904), NE * labelscalefactor); dot((10.084044692813741,-2.503922159308081),linewidth(4pt) + dotstyle); label("$G$", (10.186911369372824,-2.321884622129298), NE * labelscalefactor); dot((6.80847719440843,-2.7539040668014776),linewidth(4pt) + dotstyle); label("$K$", (6.898281450271096,-2.560191138006235), NE * labelscalefactor); dot((9.08,-7.74),linewidth(4pt) + dotstyle); label("$J$", (9.18602400268969,-7.54079731983421), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Again trivialised by isogonal line lemma. Clearly we have that $\angle BAD=30$ as $AD$ is angle bisector $\angle PAD=90$ as $AD$ is an altitude. These two conditions imply $\angle PAB=60$. Now $\angle DAC=30$ and $\angle DAQ=90\implies \angle CAQ=60$. Now we can obtain that $AP,AQ$ are isogonal lines w.r.t $\angle BAC$. Now by isogonal line lemma we just have to prove that $PB,QC$ and the $A$-median concur at a point.We claim that this is the point at infinty along the $A$- median. Let $PD \cap AB=F$ and $QD \cap AC=G$ . Now as $PDQ$ is regular we have that $\angle BPA=60 \implies PAF$ is regular since we have already proved that $\angle PAB=60$ and similiarly we can obtain $\Delta AQG$ as regular $\implies \angle AFG=180-60=120$ and $\angle AGD=180-60=120$ $\implies AFGD$ is parallelogram $\implies FB \parallel GD, FD \parallel GC \implies \Delta BFD \sim \Delta DGC$. Now observe that since $\Delta PDQ$ is regular $\implies AD$ is perpendicular bisector $\implies PF=PA=AQ=QG$ since $\Delta PAF$ and $\Delta QAG$ are regular as proved earlier . Now observe that now we have $FD=PD-PF=QD-QG=GD \implies FG \parallel PQ$ by thales theorem $\implies \angle AFG=\angle PAF=60$ and since $\angle BAC=60\implies \Delta AFG$ is regular $\implies PF=FA=FG=FD$ and similiarly $QG=AG=GF=GD$. Now by the similiarity we had obtained earlier(i.e. $\Delta BFD \sim \Delta DGC$) we have $\frac{BF}{DG}=\frac{FD}{GC}$. Putting $FD=PF,GD=GQ$ as we obtained earlier we have $\frac{BF}{GQ}=\frac{PF}{GC}$ and since $\angle PFB=\angle CGQ$ (as we have already stated earlier that $AFGD$ is a parallelogram) $\implies \Delta PFB \sim \Delta CGQ $ by $SAS$ similiarity. Now observe that since $AFGD$ is a prallelogram $\implies GD \parallel AG \implies \angle DPC=\angle PCA$ and now since we have obtained that $\Delta PFB \sim \Delta CGQ$ we have $\angle BPF=\angle QCG$. These two conditions imply $\angle BPC=\angle DPC+\angle BPF=\angle PCA+\angle QCG=\angle PCQ\implies PB \parallel QC \implies PBQC$ is a trapezium. Now let $AJ$ be the $A$-median with $J$ as midpoint of $BC$. Clearly we have that since $\Delta PDQ$ regular $\implies $ $A$ is midpoint of $PQ$ and since $J$ is midpoint of $BC \implies AJ$ is midline of trapezium $PBCQ$ and from a well known fact we have that $AJ \parallel PB \parallel QC \implies $ $AJ,PB,QC$ concur at the point at infinity as desired along the $A$-median as desired. $\blacksquare$

splendid solution(more diagrame)

Nice and Easy! Sharygin 2020 CR P11 wrote: Let $H$ be the orthocenter of a nonisosceles triangle $ABC$. The bisector of angle $BHC$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. The perpendiculars to $AB$ and $AC$ from $P$ and $Q$ meet at $K$. Prove that $KH$ bisects the segment $BC$.

Let $AB\cap QD=X$ and $AC\cap PD=Y$. Notice that after applying Pappu's Theorem on $\overline{QAP}$ and $\overline{CDB}$ we get that $\overline{X-K-Y}$ (Note that this is not that important, but this observation is quite obvious, also it makes the diagram a bit clearer.) Claim:- $QB\|PC$

Isogonal Line Lemma:- Consider two arbitrary points $P,Q$ in the plain of $\triangle ABC$. Let $P',Q'$ be the isogonal conjugate of $P,Q$ WRT $\triangle ABC$ respectively. Let $PQ\cap P'Q'=S, P'Q\cap PQ'=T$ prove that $T$ is isogonal conjugate of $S$ WRT $\triangle ABC$.

Let $M$ be the midpoint of $BC$. So, $BQ\|AM\|PC$. So, in Projective terms they all concur at Point at Infinity $P_{\infty}$. So, we need to prove that $AK,AM$ are isogonal. Notice that $\{AQ,AP\}$ and $\{AB,AC\}$ are isogonal pairs and $PB\cap QC=K$ and $QB\cap PC=P_{\infty}$ So, by Isogonal Line Lemma we get that $AK,AP_{\infty}$ are isogonal, now as $BQ\|MA\|CP$ we get that $M\in AP_{\infty}$. So, $AK,AM$ are isogonal. Hence $AK$ is the $A-\text{Symmedian}$ of $\triangle ABC$. $\blacksquare$

here is my solution-

geodomstain wrote: here is my solution- This solution is really nice! Congratulaions!

Assume that $P,B$ lie on the different side of $AD$ first. The main point is to prove the following claim (which will make our assumption valid). Claim: $BQ\parallel CP\parallel AM$ where $M$ is midpoint of $BC$. Proof: Let $B_1 = AB\cap CP$. Notice that $DP\parallel AC$ so $$\frac{CP}{PB_1} = \frac{CD}{DB} = \frac{AC}{AB}$$But since $AP$ bisects $\angle B_1AC$, we get $CP : PB_1 = CA : AB_1$ thus $AB_1 = AB$. Then notice that $CB_1\parallel AM$ so we are done. $\blacksquare$ To finish, we use isogonality lemma which is mentioned by many others. Applying this on $A$ and $BPQC$, we find $(AB, AC)$, $(AP, AQ)$ are isogonal so $(AM,AK)$ are isogonal hence we are done.

Desargues theorem Solution: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20.999cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 33.88, ymin = -11.04, ymax = 6.94; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); draw((14.68,4.28)--(10.16,-4.24)--(18.288179144513833,-1.4823552354217853)--cycle, linewidth(2.) + zzttqq); draw((14.927473024638273,-2.622540936547239)--(10.694816132192043,4.137121382607921)--(18.665183867807954,4.4228786173920795)--cycle, linewidth(2.) + green); /* draw figures */ draw((14.68,4.28)--(10.16,-4.24), linewidth(2.)); draw((14.68,4.28)--(10.16,-4.24), linewidth(2.) + zzttqq); draw((10.16,-4.24)--(18.288179144513833,-1.4823552354217853), linewidth(2.) + zzttqq); draw((18.288179144513833,-1.4823552354217853)--(14.68,4.28), linewidth(2.) + zzttqq); draw((xmin, -27.89209428638339*xmin + 413.7359441241082)--(xmax, -27.89209428638339*xmax + 413.7359441241082), linewidth(2.)); /* line */ draw((14.927473024638273,-2.622540936547239)--(10.694816132192043,4.137121382607921), linewidth(2.) + green); draw((10.694816132192043,4.137121382607921)--(18.665183867807954,4.4228786173920795), linewidth(2.) + green); draw((18.665183867807954,4.4228786173920795)--(14.927473024638273,-2.622540936547239), linewidth(2.) + green); draw((xmin, 15.663554029817188*xmin-163.38170894294262)--(xmax, 15.663554029817188*xmax-163.38170894294262), linewidth(2.)); /* line */ draw((xmin, 15.663554029817513*xmin-287.94023737249603)--(xmax, 15.663554029817513*xmax-287.94023737249603), linewidth(2.)); /* line */ draw((10.16,-4.24)--(18.665183867807954,4.4228786173920795), linewidth(2.)); draw((18.288179144513833,-1.4823552354217853)--(10.694816132192043,4.137121382607921), linewidth(2.)); draw((15.14860002202724,0.8410937345061094)--(14.927473024638273,-2.622540936547239), linewidth(2.)); draw((14.224089572256917,-2.8611776177108927)--(14.68,4.28), linewidth(2.)); draw((12.81114457841516,0.7572902230303398)--(16.796328446223114,0.900168840422424), linewidth(2.)); draw((15.14860002202724,0.8410937345061094)--(14.68,4.28), linewidth(2.)); /* dots and labels */ dot((14.68,4.28),linewidth(3.pt) + ududff); label("$A$", (14.76,4.4), NE * labelscalefactor,ududff); dot((10.16,-4.24),linewidth(3.pt) + ududff); label("$B$", (10.24,-4.12), NE * labelscalefactor,ududff); dot((18.288179144513833,-1.4823552354217853),linewidth(3.pt) + xdxdff); label("$C$", (18.36,-1.36), NE * labelscalefactor,xdxdff); dot((14.927473024638273,-2.622540936547239),linewidth(3.pt) + uuuuuu); label("$D$", (15.,-2.5), NE * labelscalefactor,uuuuuu); dot((10.694816132192043,4.137121382607921),linewidth(3.pt) + uuuuuu); label("$Q$", (10.78,4.26), NE * labelscalefactor,uuuuuu); dot((18.665183867807954,4.4228786173920795),linewidth(4.pt) + uuuuuu); label("$P$", (18.74,4.58), NE * labelscalefactor,uuuuuu); dot((14.224089572256917,-2.8611776177108927),linewidth(3.pt) + uuuuuu); label("$M$", (13.74,-3.44), NE * labelscalefactor,uuuuuu); dot((15.14860002202724,0.8410937345061094),linewidth(3.pt) + uuuuuu); label("$K$", (15.28,0.32), NE * labelscalefactor,uuuuuu); dot((16.796328446223114,0.900168840422424),linewidth(3.pt) + uuuuuu); label("$F$", (17.08,0.82), NE * labelscalefactor,uuuuuu); dot((12.81114457841516,0.7572902230303398),linewidth(3.pt) + uuuuuu); label("$E$", (12.36,0.48), NE * labelscalefactor,uuuuuu); dot((14.458873002611032,0.8163653289466544),linewidth(3.pt) + uuuuuu); label("$K'$", (13.92,0.34), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.8181818181818181) * currentpicture; /* end of picture */ [/asy][/asy] Let $PD\cap AC=F$ and $QD\cap AB=E$. Let $M$ be the midpoint of $BC$. In $\triangle PDQ$, as $AD$ is the altitude of equilateral triangle , it is also the angle bisector of $\angle PDQ$ . $\implies \angle PDA=\angle FDA=30^{0}$ ; $\angle FAD=30^{0}$. $\implies F$ lies on perpendicular bisector of $AD$. Similarly $E$ lies on perpendicular bisector of $AD$. So $FE$ is the perpendicular bisector of $AD$ ----(1). As $\triangle PDQ$ is equilateral and $AD$ is the altitude , $FE$ is the $D$-medial line, and $FE\parallel PQ$. $\implies AF=FE=AE=AP=AQ=DF=FE=x$ ----(as all sides of $\triangle PDQ$ are equal ) $\implies \triangle AFP , \triangle AEQ$ are equilateral triangles. $\implies \angle PFC=\angle QEB=120^{0}$ . This also implies that $AEDF$ is a rhombus , which also means that $PD\parallel AB$ and $DQ\parallel AC$. Let $\angle ABC=B$ and $\angle ACB=C$. Also, as $PD\parallel AB$ and $DQ\parallel AC$, and $\angle PFC=\angle QEB=120^{0}$ , $\implies \angle FDC=B; \angle FCD=C$ and $\angle EDB=C;\angle EBD=B \implies \triangle EBD \sim \triangle FDC$. $\implies \frac{EB}{ED}=\frac{FD}{FC}\implies \frac{EB}{EQ}=\frac{FP}{FC}$. As $\angle QEB=\angle CFP$; we get that $\triangle QEB \sim \triangle CFP \implies \angle BQE=\angle PCF$. As $DQ\parallel CA , \angle DQC=\angle ACQ \implies \angle BQC=\angle PCQ \implies BQ\parallel CP$. Now applying Desargues Theorem on $\triangle PFC$ and $\triangle BEQ$. We see that $PF\cap BE$ ; $FC\cap EQ$ and $PC\cap BQ$ are points at infinity in their respective directions . They are collinear(as they lie on line at infinity). So, $PB,QC,FE$ are concurrent . $\implies F,K,E$ are collinear (also known as Pappus Theorem). Let $ AM\cap FE= K'$ As $A$ is the midpoint of $PQ$, and $M$ is the point of $BC$ , and $BQ\parallel PC$ , so $AM\parallel PC\parallel QB$. We claim that $KD\parallel AM$. Proof: So, we need to prove $\frac{BD}{CD}=\frac{BK}{KP}$ As $AC\parallel QD$ and $QB\parallel PC$,and $\angle QEB=\angle CFP=120^\circ$, so $\triangle QEB\sim \triangle CFP$ (as all 3 corresponding angles are equal due to the parallel condition). $\implies EB \cdot FC = x^{2}$ By angle bisector theorem, $\frac{BD}{CD}=\frac{AB}{AC}=\frac{x+BE}{x+CF}= \frac{\sqrt{EB \cdot FC}+\sqrt{BE^{2}}}{\sqrt{EB \cdot FC}+\sqrt{CF^{2}}}=\frac{\sqrt{BE}(\sqrt{BE}+\sqrt{FC})}{\sqrt{FC}(\sqrt{BE}+\sqrt{FC})}=\sqrt\frac{BE}{FC}$. So it remains to prove that $\frac{BK}{KP}=\sqrt\frac{BE}{FC}$ or $(\frac{BK}{KP})^{2}=\frac{BE}{FC}$ Now that $BQ\parallel CP$, $\triangle BKQ \sim \triangle PKC$. $\frac{BK}{KP}=\frac{BQ}{CP}=\frac{BE}{FP}$---(as $\triangle QEB\sim \triangle CFP$). So, $(\frac{BK}{KP})^{2}= (\frac{BE}{FP})^{2}$ So , we need to prove $(\frac{BE}{FP})^{2}=\frac{BE}{FC}$ which is same as proving $BE\cdot FC= FP^{2}= x^{2}$ ; which is true . $\implies KD \parallel PC \parallel QB \parallel AM$ . So , $AK'\parallel KD$ , so $\angle K'AD=\angle KDA=\angle KAD$ ---(from 1). $\implies AD$ bisects $\angle K'AK$. As $AD$ bisects $\angle BAC$ also , so $AK'$ and $AK$ are isogonal with respect to $\angle BAC$. $\implies AK$ is isogonal to the $A$-median $\implies AK$ is the symmedian . $\blacksquare$

Attached my solution

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -6.4, xmax = 13.32, ymin = -6.52, ymax = 5.52; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); draw(arc((6.661180690074374,-4.368608991431707),0.6,134.51488163334483,163.24926101858705)--(6.661180690074374,-4.368608991431707)--cycle, linewidth(1.6) + dtsfsf); draw(arc((-1.1404752945558383,-2.020467313385908),0.6,-45.48511836665517,-16.75073898141297)--(-1.1404752945558383,-2.020467313385908)--cycle, linewidth(1.6) + dtsfsf); draw(arc((-2.9,-4.36),0.6,392.6280471403512,434.5148816333448)--(-2.9,-4.36)--cycle, linewidth(1.6) + sexdts); draw(arc((2.040710174626791,-1.1968752294626603),0.6,-147.37195285964881,-105.48511836665516)--(2.040710174626791,-1.1968752294626603)--cycle, linewidth(1.6) + sexdts); /* draw figures */ draw((-0.8,3.22)--(-2.9,-4.36), linewidth(0.8)); draw((-2.9,-4.36)--(6.661180690074374,-4.368608991431707), linewidth(0.8)); draw((-0.8,3.22)--(6.661180690074374,-4.368608991431707), linewidth(0.8)); draw((-2.9,-4.36)--(2.040710174626791,-1.1968752294626603), linewidth(0.8)); draw((6.661180690074374,-4.368608991431707)--(-1.1404752945558383,-2.020467313385908), linewidth(0.8)); draw((-0.8,3.22)--(1.1633691070687746,-4.363658701886361), linewidth(0.8)); draw((1.1633691070687746,-4.363658701886361)--(-1.1404752945558383,-2.020467313385908), linewidth(0.8)); draw((1.1633691070687746,-4.363658701886361)--(2.040710174626791,-1.1968752294626603), linewidth(0.8)); draw((-1.1404752945558383,-2.020467313385908)--(2.040710174626791,-1.1968752294626603), linewidth(0.8)); draw((-0.8,3.22)--(0.477724043024295,-4.3630413395884515), linewidth(0.8)); /* dots and labels */ dot((-0.8,3.22),linewidth(4pt) + dotstyle); label("$A$", (-0.72,3.38), NE * labelscalefactor); dot((-2.9,-4.36),linewidth(4pt) + dotstyle); label("$B$", (-3.22,-4.58), NE * labelscalefactor); dot((6.661180690074374,-4.368608991431707),linewidth(4pt) + dotstyle); label("$C$", (6.78,-4.6), NE * labelscalefactor); dot((1.1633691070687746,-4.363658701886361),linewidth(4pt) + dotstyle); label("$D$", (1.08,-4.78), NE * labelscalefactor); dot((-1.1404752945558383,-2.020467313385908),linewidth(4pt) + dotstyle); label("$Q$", (-1.52,-1.98), NE * labelscalefactor); dot((2.040710174626791,-1.1968752294626603),linewidth(4pt) + dotstyle); label("$P$", (2.14,-1.16), NE * labelscalefactor); dot((0.14836895124300128,-2.4083835322030347),linewidth(4pt) + dotstyle); label("$K$", (0.18,-2.22), NE * labelscalefactor); dot((0.477724043024295,-4.3630413395884515),linewidth(4pt) + dotstyle); label("$S$", (0.16,-4.74), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $AS$ be the $A$-symmedian. We show that $AS, BP, CQ$ are concurrent at $K$. We begin by observing that $DQ \parallel AC$ (since $\angle QDA=30 ^{\circ}= \angle DAC$) and similarly, $DP \parallel AB$. This gives us $\angle QCA= \angle CQD$ and $\angle PBA=\angle BPD$. Now we use trig ceva $$\frac{\sin \angle ACQ}{\sin \angle BCQ}=\frac{\sin \angle DQC}{\sin \angle QCD}=\frac{DC}{QD} $$Similarly, $$\frac{\sin \angle CBP}{\sin \angle ABP}=\frac{\sin \angle DBP}{\sin \angle BPD}=\frac{PD}{BD}$$$$\frac{\sin \angle BAS}{\sin \angle CAS}=\frac{AB}{AC}$$Multiplying, $$\frac{\sin \angle ACQ}{\sin \angle BCQ} \cdot \frac{\sin \angle CBP}{\sin \angle ABP} \cdot \frac{\sin \angle BAS}{\sin \angle CAS}=\frac{CD}{BD} \cdot \frac{AB}{AC}=1$$

Problem 11. Let $ABC$ be a triangle with $\angle A = 60^{\circ}$, $AD$ be its bisector, and $PDQ$ be a regular triangle with altitude $DA$. The lines $PB$ and $QC$ meet at point $K$. Prove that $AK$ is a symmedian of $ABC$.

Solution. Claim: $BQ||PC$.

Proof: As $AG$ is the midpoint of $PQ$ and $BQ||CP$, we must have $AG||BQ||PC$ where $G$ is the midpoint of $BC$. Let $P_{\infty}$ be the point of infinity for lines $BQ,AG,PC$ (The point of infinity will be same because all three lines are parallel). Now, observe that $\angle DAQ=\angle DAP=90$ and $\angle DAB=\angle DAC=30$. Hence, $AB,AC$ are isogonal and $AQ,AP$ are isogonal. Thus, by isogonal line lemma, $PB\cap QC$ and $BQ\cap CP$ forms isogonal lines when joined with $A$. More precisely, as $PB\cap QC=K, BQ\cap CP=P_{\infty}$, we have $AK, AP_{\infty}$ are isogonal with respect to $\triangle ABC$. But, we also know that $P_{\infty},A,G$ are collinear and hence $AK, AG$ are isogonal lines. Hence, as $AG$ is median line and is isogonal to $AK$, we have $AK$ is symmedian line. Feedback: The observation that $BQ||PC$ was really important to note. After this point, the problem becomes trivial.

@above ewww cartesian coordinates... This was a cool problem.

amar_04 wrote: Nice and Easy! Sharygin 2020 CR P11 wrote: Let $H$ be the orthocenter of a nonisosceles triangle $ABC$. The bisector of angle $BHC$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. The perpendiculars to $AB$ and $AC$ from $P$ and $Q$ meet at $K$. Prove that $KH$ bisects the segment $BC$.

Let $AB\cap QD=X$ and $AC\cap PD=Y$. Notice that after applying Pappu's Theorem on $\overline{QAP}$ and $\overline{CDB}$ we get that $\overline{X-K-Y}$ (Note that this is not that important, but this observation is quite obvious, also it makes the diagram a bit clearer.) Claim:- $QB\|PC$

Isogonal Line Lemma:- Consider two arbitrary points $P,Q$ in the plain of $\triangle ABC$. Let $P',Q'$ be the isogonal conjugate of $P,Q$ WRT $\triangle ABC$ respectively. Let $PQ\cap P'Q'=S, P'Q\cap PQ'=T$ prove that $T$ is isogonal conjugate of $S$ WRT $\triangle ABC$.

Let $M$ be the midpoint of $BC$. So, $BQ\|AM\|PC$. So, in Projective terms they all concur at Point at Infinity $P_{\infty}$. So, we need to prove that $AK,AM$ are isogonal. Notice that $\{AQ,AP\}$ and $\{AB,AC\}$ are isogonal pairs and $PB\cap QC=K$ and $QB\cap PC=P_{\infty}$ So, by Isogonal Line Lemma we get that $AK,AP_{\infty}$ are isogonal, now as $BQ\|MA\|CP$ we get that $M\in AP_{\infty}$. So, $AK,AM$ are isogonal. Hence $AK$ is the $A-\text{Symmedian}$ of $\triangle ABC$. $\blacksquare$ Cool solution My solution was similar

One-liner proof of $BQ \parallel CP$: Apply Pappus to the hexagon $ABQDPC$. Since $AB \parallel DP$ and $AC \parallel DQ$, this yields $BQ \parallel CP$.

Applying Pappus Theorem on $\left( \begin{array}{l} \begin{array}{*{20}{c}} P&Q&A \end{array}\\ \begin{array}{*{20}{c}} C&B&D \end{array} \end{array} \right)$ implies that $M=QD \cap AB$, $N=PD \cap AC$ and $K$ are collinear Also notice that $DM \parallel AC$, $DN \parallel AB$ and $AQ=AP=AM=AN=MN=MD=DN$ then \[\frac{{d(K,AM)}}{{d(K,AN)}} = \frac{{KM}}{{KN}} = \frac{{MB}}{{MD}} = \frac{{AB}}{{AC}}\]which means $AK$ is the symmedian of $\triangle ABC$ (Q.E.D)

Why are all these solutions so complicated? Let $PD \cap AB = X$ and $QD \cap AC = Y$, it is clear that $X,Y$ are the midpoints of $DQ,DP$ By Pappu's theorem (yay first time applying pappu's theorem) we get $K \in XY$. Now because $BX \parallel PY$ and by applying sine rule in $\triangle AXK$ and $\triangle AKY$ we get that $$\frac{\sin \angle XAK}{\sin \angle YAK}=\frac{XK}{XY}=\frac{XB}{YP}=\frac{XB}{XD}=\frac{AB}{AC}$$which is enough to show that $AK$ is the symmedian since the symmedian must satisfy the above condition $\blacksquare$