Note that lines $BQ$ and $CP$ are reflections of $BC$ over lines $BH$ and $CH$, respectively, so $H$ is the incenter of triangle $BSC$. In addition, we have $$\measuredangle PTQ = 2\measuredangle PAQ = 2\measuredangle BAC = \measuredangle CSB = \measuredangle PSQ$$ so quadrilateral $TPSQ$ is cyclic. Since $\angle PAQ$ is acute, $T$ and $S$ lie on opposite sides of line $PQ$. Hence, since $TP = TQ$, $T$ is the midpoint of arc $\overset{\Huge\frown}{EF}$ of $(SPQ)$ not containing $S$. It follows that both $T$ and $H$ lie on the interior angle bisector of $\angle BSC$, as desired.

[asy][asy] size(7cm); import olympiad; import geometry; pair A,B,C,P,Q,S,H,T; A=(-4.2,9.3); B=(-6,-1); C=(0,-1); P=intersectionpoints(circle(C,length(segment(B,C))),A--B)[0]; Q=intersectionpoints(circle(B,length(segment(B,C))),A--C)[1]; S=extension(B,Q,C,P); H=orthocenter(A,B,C); T=circumcenter(A,P,Q); draw(A--B--C--cycle); draw(B--H); draw(C--H); draw(P--H); draw(Q--H); draw(P--T); draw(Q--T); draw(C--P); draw(B--Q); draw(P--Q); draw(T--H,dashed); draw(circle(A,P,Q)); draw(circle(B,P,H),dotted); draw(circle(C,Q,H),dotted); draw(circle(S,P,Q),dotted); dot(A^^B^^C^^P^^Q^^S^^H^^T); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,SW); label("$Q$",Q,NE); label("$S$",S,W); label("$H$",H,SE); label("$T$",T,NW); [/asy][/asy] It suffices to prove both $\overline{SH}$ and $\overline{TH}$ bisect $\angle BSC.$ Claim: $BHSP$ is cyclic. Proof. Notice $\overline{CH}$ bisects $\angle PCB$ as $PC=CB$ and $\overline{CH}\perp\overline{PB}.$ Similarly, $\overline{BH}$ bisects $\angle CBQ$ so $H$ is the incenter of $\triangle SBC.$ Hence, $$\angle BPH=\angle HBP=90-\angle A=\angle BHC-90=\angle BSH.$$$\blacksquare$ Similarly, $CHSQ$ is cyclic. Claim: $SPTQ$ is cyclic. Proof. Note \begin{align*}\angle PTQ&=2\angle A\\&=180-\angle AHC-\angle HBA\\&=180-\angle BPH-\angle HQC\\&=180-\angle BSH-\angle HSC\\&=180-\angle QSP.\end{align*}$\blacksquare$ Thus, $$\angle TSP=\angle TQP=\angle QPT=\angle QST$$and $\overline{TS}$ bisects $\angle QSP.$ $\square$

If you want circles, I'll construct some circles. [asy][asy] import olympiad; size(250); pair C = origin, A = (5, 0), B = (1, 3.5); pair H = orthocenter(A, B, C); pair P = 2*foot(B, C, H) - B, Q = 2*foot(C, B, H) - C; pair S = extension(B, Q, C, P); pair T = circumcenter(A, P, Q); draw(A--B--C--cycle, green); draw(C--P, yellow); draw(B--Q, yellow); draw(circumcircle(A, P, Q), lightgreen); draw(B--H, darkgreen); draw(C--H, darkgreen); draw(circumcircle(C, S, Q), orange); draw(circumcircle(B, S, P), orange); draw(circumcircle(A, B, C), darkolive); draw(circumcircle(P, S, Q), orange); draw(H--S--T, dashed); dot("$A$", A, SE); dot("$B$", B, N); dot("$C$", C, SW); dot("$H$", H, W); dot("$S$", S, NE); dot("$T$", T, N); dot("$P$", P, NE); dot("$Q$", Q, dir(270)); [/asy][/asy] Claim. $\overline{SH}$ bisects $\angle BSC$. It suffices to show that $HSQC$ is cyclic by Fact 5. Verify that $$\angle CSQ = \angle CBS+\angle SCB = 2\angle A,$$and $$\angle CHQ = 180^\circ-2(C-90^\circ+B) = 2\angle A.$$The claim follows. $\blacksquare$ Claim. $\overline{ST}$ bisects $\angle PSQ$. Again, it suffices to show $SPTQ$ is cyclic. This follows because $$\angle PTQ = 2 \angle A = \angle CSQ. \ \blacksquare$$From here one can verify $$\angle HST = \angle HSB + \angle BSP + \angle PST = \frac 12 \cdot 360^\circ = 180^\circ.$$

luminescent wrote: Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear. O. The first thing is to realice the paper of $H$ on the diagram. Claim 1: $H$ is the incenter of $\triangle BSC$ Proof: This is just becuase $P$ is the reflection of $B$ over $CH$ and $Q$ is the reflection of $C$ over $BH$. Claim 2: $TPQS$ is cyclic. Proof: Clearly it is enough to show that $\angle PSQ=180-2\angle BAC$ becuase $T$ is the circumcenter of $\triangle APQ$. Now by angle chasing using the reflections and the ortocenter we get. $$\angle PSQ=\angle CPB+\angle PBS=2\angle ABC-\angle SBC=2\angle ABC+2\angle ACB-180=180-2\angle ABC$$So the claim is proven as $\angle PTQ=2\angle ABC$ Finishing: Now we have $TP=TQ$ so $TS$ bisects $\angle BSC$ using Claim 2 and becuase of Claim 1 we have that $SH$ bisects $\angle BSC$ so we are done.

We prove that line $TH$ is the internal angle bisector of $\angle BSC=\angle QSP$. Let $A, B, C$ denote $\angle CAB, \angle ABC, \angle BCA $ respectively. Claim 1 : $ST$ bisects $\angle PSQ$. Proof : We first prove that $\square TPSQ$ is cyclic. As, $BQ=BC$, $\angle QBC=180-2\angle QCB=180-2C$. Hence, $\angle SBC=\angle QBC= 180-2C$. As, $CP=CB$, $\angle PCB=180-2\angle PCB=180-2B$. Hence, $\angle SCB=\angle PCB= 180-2B$. Hence, $\angle BSC=180-2A$. Hence, $\angle PSQ=180-2A $. As, $T$ is circumcenter of $\triangle APQ$, $\angle PTQ=2\angle PAQ= 2A$. Hence, $\angle PSQ+\angle PTQ= (180-2A)+(2A)=180$. Hence, $\square TPSQ$ is cyclic Now, note that $PT=QT$. Hence, from $\angle PST= \angle QST$, from cyclic $\square TPSQ$. Hence proved claim 1 ! Claim 2 : $SH$ bisects $\angle BSC$. Proof :We first prove that $\square PBHS$ is cyclic. As $CH \perp BP$ in isosceles $\triangle CPB$, $CH$ is also the perpendicular bisector of $PB$. Hence, $HP=HB \implies \angle HPB=\angle HBP$. But, $\angle HBP =\angle ABH=90-A$. Hence, $\angle BHP=2A$. But, $\angle BSC=180-2A$(Proved in claim 1). Hence, $\angle PSB=2A$. Hence, $\angle PHB=\angle PSB \implies \square PBHS$ is cyclic. Hence, $\angle BSH=\angle BPH=90-A$. But $\angle BSC=180-2A$. Hence, $SH$ bisects $\angle BSC$. Hence proved claim 2! But, internal angle bisectors of $\angle PSQ$ and $\angle BSC$ must be same; as $P-S-C$ and $Q-S-B$. Hence, $ST \equiv SH$ from claim 1 and 2 Hence, $T-S-H$. Hence proved !

Lmao this problem is pretty funny We claim $T$ and $H$ lie on the internal angle bisector of $\angle BSC$. For $H$, recall that \[ \angle CBH = 90^{\circ}- \angle C = 90^{\circ}- \angle BQC = \frac{1}{2}\angle CBQ\]Which implies that $H$ lies on the angle bisector of $\angle CBS$. By symmetry, it also lies on the angle bisector of $\angle SCB$. Thus, $H$ is the incenter of $BSC$, implying that $H$ lies on the angle bisector of $\angle BSC$. Now, for $T$. Notice that $PT=PQ$. So, to prove that it lies on the angle bisector of $\angle BSC$, it is sufficient to show $SPTQ$ is cyclic. Indeed, \[ \angle PTQ = 2 \angle PAQ \]Notice that $\angle APS = 180^{\circ}-\angle B$ and $\angle AQS = 180^{\circ}-\angle C$. Thus, \[ 2\angle PAQ = 180^{\circ}-\angle PTQ \]As desired. $\blacksquare$

Funny indeed Time to ditch dangles, write $\angle BCP = 2\theta_1$ and $\angle CBQ = 2\theta_2$ then $\angle PSQ = \angle BSC = \pi - 2\theta_1 - 2\theta_2$ and $\angle BAC = \theta_1 + \theta_2$ so $\angle PTQ = 2\theta_1 + 2\theta_2$ which means $PTQS$ is cyclic and since $TP=TQ$ hence $TS$ bisects $\angle PSQ$. Since $CH$ is the perpendicular bisector of $\overline{BP}$ and $BH$ is the perpendicular bisector of $\overline{CQ}$ hence $H$ lies on the angle bisector of $\angle BCS$ and $\angle CBS$ implying it is the incenter of $\triangle BSC$ so $HS$ bisects $\angle BSC$ but since $\angle BSC$ and $\angle PSQ$ are vertically opposite angles it means $T-S-H$.

10 minute solve Also yes where are the circles? First note that $H$ is the incenter in $\triangle SBC$ and $$\angle PSQ = 180-2 \cdot \angle BAC = 180-\angle PTQ \implies \odot(TPSQ)$$This means $T$ is the midpoint of the minor arc $PQ$ which means both $H$ and $T$ lie on the angle bisector of $\angle BSC$ which implies $T-H-S \ \blacksquare$

Claim-1- $\square PSHB$ (by symmetry $QSHC$) is cyclic Proof- Let $D=HC \cap AB$ Note that since in isosceles $\triangle PBC$; $CD$ is an altitude, it is also the perpendicular bisector of $PB$. Further, notice that $\angle PBH=\angle BPH=90-A \implies \angle PHB=2A$ Lastly, $\angle PSB= 180-(\angle PSQ)=360-180-(\angle B)+180-(\angle C)-\angle A=2A$ ...... ($\star$) We have proved the claim. As a corollary, we have $\angle BSH=\angle BPH=90-A$, but due to symmetry (and trivial angle chase) we have $\angle CSH=90-A$ hence we have $SH$ bisects $\angle BSC$ Claim-2- $\square PTQS$ is cyclic Proof Firstly note that $\angle PTQ=2\angle PAQ=2A$ The claim is true due to ($\star$) NOTE that it suffices to show that $ST$ bisects $\angle PSQ$, but it is trivial since angles subtended in equal arcs are equal ($TP=TQ$)

In the following solution, points $S$ and $T$ are switched because I'm an idiot. Claim: $SPTQ$ is cyclic. Proof. First, we have $\angle PSQ=2\angle PAQ=2\angle A$. But we also have \begin{align*} \angle PTQ=360^\circ-\angle TPA-\angle PAQ-\angle APT=360^\circ-(180^\circ-\angle B)-\angle A-(180^\circ-\angle C)=\angle B+\angle C-\angle A=180^\circ-2\angle A, \end{align*}as required. Note that this also implies that $\angle PTS=\angle STQ=90^\circ-\angle A$, since $SP=SQ$. Claim: $PTHB$ is cyclic. Proof. We have $\angle PHB=180^\circ-2(90^\circ-\angle A)=2\angle A$. But we also have $\angle BTP=180^\circ-\angle TPB-\angle PBT=180^\circ-\angle B-(\angle C-\angle A)=2\angle A$, so indeed, $PTHB$ is cyclic. This implies that $\angle PTH=180^\circ-\angle PBH=180^\circ-(90^\circ-\angle A)=90^\circ+\angle A$. Thus, as $\angle SPT+\angle PTH=(90^\circ-\angle A)+(90^\circ+\angle A)=180^\circ$, $S, T$< and $H$ are collinear as desired. Remark: Indeed, where are the circles? Definitely not the circles centered at $B$ and $C$ with radius $BC$, which are a clever decoy.

Let $D, E, F$ be the feet of the altitudes from $A, B, C$ in $\triangle ABC$. Claim 1: $H$ is the Miquel point of quadrilateral $APSQ$. Proof: $\angle HAP = \angle DAB = 90 - \hat{B}$, and $\angle HCP = \angle FCP = \angle BCF = 90 - \hat{B}$ by symmetry. Hence, by converse of angles in the same segment, $APHC$ is a cyclic quad, and similarly, $ABHQ$ is a cyclic quad. Thus, $H$ is the Miquel point of $APSQ$. $\square$ Claim 2: $\angle PST = \angle TSQ = 90 - \hat{A}$. Proof: Note that because of Claim 1, $PBHS$ and $QSHC$ are cyclic. $\angle PST = \angle CSH = 180 - \angle SHC - \angle HCS = 180 - (180 - \angle CQS) - \angle BCF = \angle CQS - (90 - \hat{B}) = \hat{C} + \hat{B} - 90 = 90 - \hat{A}$. Similarly, $\angle TSQ = 90 - \hat{A}$. $\square$. Claim 3: $TPSQ$ cyclic. Proof: $\angle QTP = 2 \angle QAP = 2 \hat{A}$, and $\angle PSQ = \angle PST + \angle TSQ = 180 - 2 \hat{A}$, so by converse of opposite angles in a cyclic quad, $TPSQ$ cyclic. $\square$ Claim 4: $T, S, H$ collinear. Proof: By Claim 2, $\overline{SH}$ is the internal angle bisector of $\angle PSQ$. But in $(TPSQ)$, $T$ is the arc-midpoint of $PQ$ not containing $S$, and hence $ST$ is also the internal angle bisector of $\angle PSQ$. Hence, $T, S, H$ are collinear. $\square$ $\blacksquare$

Let $D=BQ\cap (APQ), E=PC\cap (APQ)$. Note that $\angle SDP=\angle A$ and \begin{align*} \angle SPD&=180^\circ-\angle SPB-\angle APD\\ &=180^\circ-\angle B-\angle C\\ &=\angle A \end{align*}So $\triangle SPD$ is isosceles. From PoP $SD\cdot SQ=SP\cdot SE$. Since $SP=SD$ therefore $DQ=PE$, means $PDQE$ is an isosceles trapezoid since it is cylcic. So $S$ lies on the perpendicular bisector of $PD$ and $EQ$. We will show $H$ lies on that perpendicular bisector. Or equivalent to show $HE=HQ$. Let $F$ be the foot of the $B$-altitude. Notice that \begin{align*} \angle CHQ&=2\angle CHF\\ &=2(90^\circ-\angle HCF)\\ &=2(90^\circ-(90^\circ-\angle A))\\ &=2\angle A\\ &=2\angle CPD\\ &=2\angle CEQ %\angle CEQ=\angle CPD=\angle A \end{align*}which means $H$ is the circumcenter of $(CQE)$ and hence $HE=HQ$.

Easy problem. Just note some cyclic quadrilaterals and angle chasing

Notice that $\angle{PSQ}=\angle{BSC}=180^{\circ}-2A$ as $\angle{SBC}=180^{\circ}-2C$ and $\angle{SCB}=180^{\circ}-2B$. Also, $\angle{PTQ}=2A$ thus $\text{TPSQ}$ is cyclic. Dropping the perpendiculars from $CH$ and $BH$ to sides $AB$ and $AC$ respectively would imply that $H$ is the angle bisector of $\angle{SBC}$ and $\angle{SCB}$ and thus, also of $\angle{BSC}$. Hence, $SH$ is the internal angle bisector of $\angle{BSC}$. As $PT=QT$, $\angle{PST}=\angle{PQT}=\angle{QPT}=\angle{QST}$ and also, $\angle{PSB}=\angle{QSB}$. Hence, $TS$ is the external angle bisector of $\angle{BSC}$ and $T-S-H$ $\blacksquare$.

luminescent wrote: Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear. Firstly we start by doing some angle chasing: $$\angle PSQ=\angle SBP+\angle SPB =\angle B-\angle CBQ +\angle B =2\angle B - (180^\circ -2\angle C)=2(\angle B+\angle C)-180^\circ=180^\circ-2\angle A$$But, $\angle PTQ=2\angle A\implies PTQS$ cyclic. But we also have that $TP=TQ \implies ST$ is the angle bisector of $\angle PSQ.$ Also since $CB=CP$, and $CH\perp PB\implies \angle BCH=\angle PCH$, similarly $\angle CBH=\angle SBH\implies H$ is the incenter of $\triangle SBC\implies SH$ is the angle bisector of $\angle BSC\implies \overline{T-S-H}$ are collinear.$\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -10.646666666666658, xmax = 7.753333333333355, ymin = -3.346666666666663, ymax = 7.826666666666668; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ubqqys = rgb(0.29411764705882354,0.,0.5098039215686274); draw((-2.98,5.533333333333335)--(-5.433333333333317,-1.573333333333332)--(-0.633333333333317,-1.45333333333333)--cycle, linewidth(0.8)); /* draw figures */ draw((-2.98,5.533333333333335)--(-5.433333333333317,-1.573333333333332), linewidth(0.8)); draw((-5.433333333333317,-1.573333333333332)--(-0.633333333333317,-1.45333333333333), linewidth(0.8)); draw((-0.633333333333317,-1.45333333333333)--(-2.98,5.533333333333335), linewidth(0.8)); draw((-1.534115655163983,1.2285413066624968)--(-5.433333333333317,-1.573333333333332), linewidth(0.8) + red); draw((-4.337059197869649,1.6022868525478613)--(-0.633333333333317,-1.45333333333333), linewidth(0.8) + red); draw(circle((-2.693029270754494,3.2345041635124936), 2.316671696952953), linewidth(0.8) + qqwuqq); draw((-2.693029270754494,3.2345041635124936)--(-2.8242417360585397,-0.6969972243250036), linewidth(0.8) + ubqqys); draw((-4.337059197869649,1.6022868525478613)--(-1.534115655163983,1.2285413066624968), linewidth(0.8)); draw((-2.693029270754494,3.2345041635124936)--(-4.337059197869649,1.6022868525478613), linewidth(0.8)); draw((-2.693029270754494,3.2345041635124936)--(-1.534115655163983,1.2285413066624968), linewidth(0.8)); draw(circle((-2.8862940403917934,1.7850949490050483), 1.4622374438790875), linewidth(0.8)); draw((-5.433333333333317,-1.573333333333332)--(-1.0837244942486495,-0.11239601333541793), linewidth(0.8) + blue); draw((-0.633333333333317,-1.45333333333333)--(-4.885196265601483,0.01447675960726485), linewidth(0.8) + blue); /* dots and labels */ dot((-2.98,5.533333333333335),dotstyle); label("$A$", (-2.926666666666652,5.666666666666669), NE * labelscalefactor); dot((-5.433333333333317,-1.573333333333332),dotstyle); label("$B$", (-5.686666666666654,-1.89333333333333), NE * labelscalefactor); dot((-0.633333333333317,-1.45333333333333),dotstyle); label("$C$", (-0.58,-1.32), NE * labelscalefactor); dot((-1.534115655163983,1.2285413066624968),linewidth(4.pt) + dotstyle); label("$Q$", (-1.366666666666651,1.12), NE * labelscalefactor); dot((-4.337059197869649,1.6022868525478613),linewidth(4.pt) + dotstyle); label("$P$", (-4.69,1.4), NE * labelscalefactor); dot((-2.693029270754494,3.2345041635124936),linewidth(4.pt) + dotstyle); label("$T$", (-2.766666666666652,3.36), NE * labelscalefactor); dot((-2.790098728363682,0.32602510687073094),linewidth(4.pt) + dotstyle); label("$S$", (-2.7,0.5733333333333362), NE * labelscalefactor); dot((-2.8242417360585397,-0.6969972243250036),linewidth(4.pt) + dotstyle); label("$H$", (-2.686666666666652,-0.5333333333333302), NE * labelscalefactor); dot((-1.0837244942486495,-0.11239601333541793),linewidth(4.pt) + dotstyle); label("$E$", (-1.0333333333333174,0.), NE * labelscalefactor); dot((-4.885196265601483,0.01447675960726485),linewidth(4.pt) + dotstyle); label("$F$", (-5.206666666666654,0.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

It suffices to prove that the projections of $T$, $S$, $H$ on the lines $AB$ and $AC$ make segments with equal ratios on $AB$ and $AC$. Let $M,D,K$ the projections on $AB$ and $N,E,L$ on $AC$. Then, it suffices $$\frac{MD}{MK}=\frac{NE}{NL}.$$But $MK=AB/2$ and $NL=AC/2$, so it suffices to show that $$\frac{MD}{AB}=\frac{NE}{AC}.$$Thales theorem gives $$\frac{MD}{MP}=\frac{CS}{CP}=\frac{CS}{CB}.$$Therefore, we have to prove that $$\frac{MP\cdot CS}{AB}=\frac{NQ\cdot BS}{AC}.$$By the law of sines we have $$\frac{PB}{BS}=\frac{\sin (2B+2C)}{\sin B}$$and $$\frac{CQ}{CS}=\frac{\sin (2B+2C)}{\sin C}.$$Therefore it suffices to prove that $\frac{AB}{AC}=\frac{\sin C}{\sin B}$, which is clear.

[asy][asy] defaultpen(fontsize(8pt)); size(5cm); pair A, B, C, S, H, P, Q, Pp; A=dir(100); B=dir(230); C=dir(-50); H=A+B+C; P=2*foot(B,C,H)-B; Q=2*foot(C,B,H)-C; S=intersectionpoint(B--Q, C--P); Pp=2*foot(P,S,H)-P; draw(A--B--C--cycle); draw(B--Q); draw(C--P); draw(circumcircle(A,P,Q),gray+dashed); draw(P--Pp); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$S$",S,dir(-90)); dot("$H$",H,dir(H)); dot("$P$",P,dir(210)); dot("$Q$",Q,dir(20)); dot("$P'$",Pp,dir(0)); [/asy][/asy] Note $H$ is the incenter of $\triangle BCS$, so if $P'$ is the reflection of $P$ across $\overline{SH}$, it suffices to show $APP'Q$ cyclic (and then the circumcenter would lie on $\overline{SH}$, which is the perpendicular bisector of $PP'$). Indeed, \begin{align*} \angle PP'Q &= 180^\circ - \angle PP'S = 180^\circ - \frac{180^\circ - \angle PSP'}{2}=180^\circ - \frac{180^\circ - \angle BSC}{2}\\ &=180^\circ - \frac{\angle SBC + \angle SCB}{2}=180^\circ - \angle HBC - \angle HCB\\ &=180^\circ - (90^\circ - \angle C) - (90^\circ - \angle B) = 180^\circ - \angle A\\ &=180^\circ - \angle PAQ. \end{align*}

I can't believe nobody has complex bashed this yet.

"People who are poor lacking circles should just get a job circles" We begin with the following key claim: Claim: $TPSQ$ is cyclic. Proof: We clearly have $\angle PTQ=2\angle A$, and (by looking at $APSQ$): $$\angle PSQ=360^\circ-\angle A-(180-\angle B)-(180-\angle C)=\angle B+\angle C-\angle A=180^\circ-2\angle A,$$hence $\angle PTQ+\angle PSQ=360^\circ$ so $TPSQ$ is cyclic as desired. Since $TP=TQ$, it follows that $\overline{ST}$ bisects $\angle PSQ$. Let $H'=(PSB) \cap \overline{ST}$, so we have $\angle PST=\angle QST=\angle BSH'=\angle CSH'=90^\circ-\angle A$. Then, $$\angle PBH'=\angle PST=\angle BSH'=\angle BPH',$$hence $H'B=H'P$ and $H'$ lies on the perpendicular from $C$ to $\overline{AB}$. Thus, $$\angle QCH'=90^\circ-\angle A=\angle TSQ=180^\circ-\angle QSH',$$so $QSH'C$ is cyclic as well. Then, $\angle CQH'=\angle CSH'=\angle TSQ=\angle QCH'$, hence we have $H'C=H'Q$ and $H'$ lies on the perpendicular from $B$ to $\overline{AC}$ as well. It thus follows that $H'=H$, hence $T,H,S$ are collinear as desired. $\blacksquare$ Edit: did not realize $H$ is the incenter of $\triangle BCS$ which instakills the problem after the claim

Let $CP$ and $BQ$ meet the circumcircle of $\triangle APQ$ again at $X$, $Y$, respectively. Then $X, Q, Y, P$ are vertices of an isosceles trapezoid by power of points. Note that $\angle BHP=2\angle A$ and $\angle BYP=\angle A$. Since $BH=HP$, we have $H$ is the circumcenter of $\triangle BYP$. Hence, $HY=HP$ and now we have a conclusion that $T,S,H$ are collinear as desired.

Let $E = BH \cap CA$, $F = CH \cap AB$, the midpoint of $AH$ be $M$, the circumcenter of $ABC$ be $O$, the $A$-Queue point be $Q_a$, and $X = AO \cap CP$. Clearly, $E$ is the midpoint of $CQ$ and $F$ is the midpoint of $BP$. It's well-known that $AEHFQ_a$ is cyclic with diameter $AH$, so $Q_a$ is the Miquel Point of $BCEF$, yielding $Q_aBF \overset{+}{\sim} Q_aCE$. Now, because $P$ is the reflection of $B$ over $F$ and $Q$ is the reflection of $C$ over $E$, we have $Q_aBFP \overset{+}{\sim} Q_aCEQ$, implying $$Q_aBC \overset{+}{\sim} Q_aFE \overset{+}{\sim} Q_aPQ.$$Thus, $AQ_aPQ$ is cyclic, so adding the centers of our three circumcircles gives $$Q_aBCO \overset{+}{\sim} Q_aFEM \overset{+}{\sim} Q_aPQT$$which implies $$Q_aBFP \overset{+}{\sim} Q_aCEQ \overset{+}{\sim} Q_aOMT.$$It follows that $M$ is the midpoint of $OT$, so $AOHT$ is a parallelogram. Now, observe $$\angle PSQ = \angle CSB = 180^{\circ} - \angle SBC - \angle SCB = 180^{\circ} - 2 \angle CBH - 2 \angle BCH$$$$= 180^{\circ} - 2(180^{\circ} - \angle BHC) = 180^{\circ} - 2 \angle A = 180^{\circ} - 2 \angle PAQ$$$$= 180^{\circ} - \angle PTQ$$so $PSQT$ is cyclic. Since $TP = TQ$, we know $T$ is the midpoint of arc $PQ$, which implies $$\angle PST = \frac{\angle PSQ}{2} = \frac{180^{\circ} - 2 \angle A}{2} = 90^{\circ} - \angle A.$$In addition, isogonality yields $$\angle PXA = \angle BPX - \angle PAX = \angle BPC - \angle BAO = \angle PBC - \angle HAC$$$$= \angle B - (90^{\circ} - \angle C) = 90^{\circ} - \angle A$$so $\angle PST = \angle PXA$. Hence, $$ST \parallel XA \equiv AO \parallel HT$$which finishes. $\blacksquare$ Remarks: Objectively, this solution sucks. Noticing that $H$ is the incenter of $BSC$ is the best way to solve this question.

Sol sketch:- Note that $ST$ bisects $\angle PSQ$ since $STPQ$ is a cyclic quad. & $TP=TQ$ and $H$ is incenter of $BSC$ so done.

İt is so easy problem i think for egmo.İ did it very easyly

A different approach, I guess: Solution. WLOG, $AC>AB$. Let $D$ be the $B$-foot of altitude on $AC$. Given that $\bigtriangleup BCP$ and $\bigtriangleup CBQ$ are isosceles at $C$ and $B$, respectively, simple angle chasing leads us to get $\angle QSC = 2\angle BAC = \angle PTQ$, so $PTQS$ is cyclic. On the other hand, $\angle QHC = 2\angle DHC = 2\angle BAC$. We conclude that $QSHC$ is inscribed. What's more, $\bigtriangleup QTP$ and $\bigtriangleup QHC$ are similar isosceles triangles, since $HD$ is the perpendicular bisector of $\overline{QC}$ and the angles corresponding to their apexes are congruent. Hence, $Q$ is the center of spiral similarity carrying $\overline{TP}$ to $\overline{HC}$, so it also sends $\overline{TH}$ to $\overline{SC}$, which gives us $\angle THQ = \angle SCQ = \angle SHQ$. Since $T$ and $S$ lie on the same side with respect to $QH$, the required assertion follows. $\Box$

Uh how do you notice that $H$ is the incenter

Hard. Claim 1: $H$ is the incenter of $BSC.$ Proof. Notice that $HB$ and $HC$ bisects $\angle CBQ$ and $\angle PCB,$ respectively. $\blacksquare$ Claim 2: $PQST$ is cyclic. Proof. Set $\angle BPC=\angle PBC= X$ and $\angle BQC=\angle BCQ= Y.$ So $\angle BCP=180^{\circ}-2X$ and $\angle QBC=180^{\circ}-2Y.$ In $\triangle CBS$ we have, $\angle PSQ=\angle CSB=2X+2Y-180^{\circ} \implies \angle BSH=\angle CSH=X+Y-90^{\circ}.$ In $\triangle ABC$ we have, $\angle CAB=180^{\circ}-X-Y \implies \angle PTQ=360^{\circ} -2X-2Y \implies \angle PTQ+\angle PSQ=180^{\circ}.$ So $PQST$ is cyclic. $\blacksquare$ Claim 3: $H,T,S$ are collinear. Proof. $PT=QT\implies \angle QPT=\angle PQT=X+Y-90^{\circ} \implies \angle PQT=\angle TSP=\angle HSC.$ So $P,S,C$ are collinear and thus $H,T,S$ are collinear. $\blacksquare$

Here is a link to a video solution https://youtu.be/w_RpeWCjMNU

Since $BH,CH$ bisect angles $CBQ,BCP$ respectively, $H$ is the incenter of $BCS.$ Observe that $$\measuredangle PSQ=\measuredangle PCB+\measuredangle CBQ=2\measuredangle BAC=\measuredangle PTQ\implies T\in \odot (PQT).$$Hence $ST$ bisects angle $PSQ\implies T\in SH.$ We are done.

Note that $H$ is incenter of $BSC$ so $\angle BSH = \frac{\angle BSC}{2}$ and $\angle BSC = \angle 180 - \angle CBQ - \angle BCP = \angle 180 - 2\angle A$ so $\angle BSH = \angle 90 - \angle A$. Note that $\angle PSQ = \angle BSC = \angle 180 - 2\angle A$ and $\angle PTQ = 2\angle A$ so $PTQS$ is cyclic so $\angle QST = \angle QPT = \angle 90 - \angle A$ so $T,H,S$ are collinear.

angle chasing gives that TPSQ is cyclic thus TS bisects angle PSQ also, SHBP and SHQC are both cyclic Thus, H is miquel point of quadilateral APSQ thus,QHC is similar to PHB hence, HS bisects angle BSC HENCE PROVED S-T-H are collinear

I just want to know why am I so bad with directed angles. Anyways here is a terrible looking angle chase... Solution: We won't really introduce any other points in the solution. It is just cyclic quadrilaterals and angle chasing. [asy][asy] import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(11); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4, xmax = 9, ymin = -0.5, ymax = 10; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); draw((0.47161707998986097,9.029775731967431)--(-1.7348933250166911,0.7343647089621917)--(5.930863521293179,0.7602625361456711)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw((0.47161707998986097,9.029775731967431)--(-1.7348933250166911,0.7343647089621917), linewidth(1) + qqwuqq); draw((-1.7348933250166911,0.7343647089621917)--(5.930863521293179,0.7602625361456711), linewidth(1) + qqwuqq); draw((5.930863521293179,0.7602625361456711)--(0.47161707998986097,9.029775731967431), linewidth(1) + qqwuqq); draw((-0.7089727971461135,4.591329279043008)--(5.930863521293179,0.7602625361456711), linewidth(1) + blue); draw((1.301056915435547,7.773363562613006)--(-1.7348933250166911,0.7343647089621917), linewidth(1) + blue); draw(circle((-1.1201722349380954,7.076941688134685), 2.519395567565953), linewidth(1) + ffxfqq); draw((1.301056915435547,7.773363562613006)--(-0.7089727971461135,4.591329279043008), linewidth(1) + blue); draw(circle((0.48163283480824953,6.065112305724049), 1.8946182149907076), linewidth(1) + linetype("4 4") + red); draw(circle((-3.3497351087582654,5.605061433960277), 5.131413116620483), linewidth(1) + linetype("4 4") + red); draw(circle((4.0249963076451305,4.536844520627115), 4.230236556218622), linewidth(1) + linetype("4 4") + red); draw((1.301056915435547,7.773363562613006)--(0.49466954880347824,2.2062449631366), linewidth(1) + blue); draw((0.49466954880347824,2.2062449631366)--(5.930863521293179,0.7602625361456711), linewidth(1) + blue); draw((-1.1201722349380934,7.076941688134685)--(0.49466954880347824,2.2062449631366), linewidth(1) + linetype("4 4")); /* dots and labels */ dot((0.47161707998986097,9.029775731967431),dotstyle); label("$A$", (0.6183605061281343,9.268641420070281), NE * labelscalefactor); dot((-1.7348933250166911,0.7343647089621917),dotstyle); label("$B$", (-2.0898709473794024,0.8749174449608395), NE * labelscalefactor); dot((5.930863521293179,0.7602625361456711),dotstyle); label("$C$", (5.85547033675198,1.054270521352067), NE * labelscalefactor); dot((-0.7089727971461135,4.591329279043008),linewidth(4.pt) + dotstyle); label("$P$", (-1.1751702577841416,4.497849588063633), NE * labelscalefactor); dot((1.301056915435547,7.773363562613006),linewidth(4.pt) + dotstyle); label("$Q$", (1.2999021964147992,8.120781731166426), NE * labelscalefactor); dot((0.49466954880347824,2.2062449631366),linewidth(4.pt) + dotstyle); label("$H$", (0.29552496862392463,1.6461356734431172), NE * labelscalefactor); dot((-1.1201722349380934,7.076941688134685),linewidth(4.pt) + dotstyle); label("$T$", (-1.4800704876492285,7.206081041571166), NE * labelscalefactor); dot((-0.1984212694812931,4.296750183359536),linewidth(4.pt) + dotstyle); label("$S$", (0.02649535403708324,4.390237742228896), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $AQHB$ and $APHC$ are cyclic quadrilaterals. Proof: For $AQHB$ observe \[\measuredangle AHB = \measuredangle BCA = \measuredangle AQB\]A similar angle chase can show $APHC$ is cyclic too. $\square$ One can also show that $TQPS$ is cyclic quadrilateral because \begin{align*} \measuredangle QSC &= \measuredangle SQC + \measuredangle QCS \\ &= \measuredangle ACB + (\measuredangle ACB - \measuredangle PCB) \\ &= 2\measuredangle ACB + 2\measuredangle CBA \\ &= 2\measuredangle CAB \\ &= 2\measuredangle QAP = \measuredangle QTP \end{align*} Claim: $SHCQ$ and $SHBP$ are cyclic quadrilaterals. Proof: We will only show $SHCQ$ cyclic as the latter one follows symmetrically. \begin{align*} \measuredangle QHC &= \measuredangle AHC - \measuredangle AHQ \\ &= \measuredangle CBA + \measuredangle QBA \\ &= \measuredangle CBA + \measuredangle CBA - \measuredangle CBQ \\ &= 2\measuredangle CBA - (2\measuredangle BCA) \\ &= 2\measuredangle ACB + 2\measuredangle CBA \\ &= 2\measuredangle ACB - \measuredangle PCB \\ &= \measuredangle ACB + \measuredangle ACP \\ &= \measuredangle BQC + \measuredangle ACP = \measuredangle QSC \end{align*}so we are done. $\square$ To show $T-S-H$ are collinear, it suffices to prove $\measuredangle TSQ = \measuredangle HSQ$. \[\measuredangle HSQ = \measuredangle HCQ = 90^\circ - \measuredangle CAB = \measuredangle TPQ = \measuredangle TSQ\]and thus we are done with the solution. $\blacksquare$

Probably the same as others, but still posting.. [asy][asy]/* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.3cm); 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 = -13.8, xmax = 5.6, ymin = -6.7, ymax = 5.1; /* image dimensions */ pen wrwrwr = rgb(0.4,0.4,0.4); draw((-6.9,3.9)--(-9.4,-4.3)--(-2.2,-4.3)--cycle, linewidth(0.4) + wrwrwr); /* draw figures */ draw((-6.9,3.9)--(-9.4,-4.3), linewidth(0.4) + wrwrwr); draw((-9.4,-4.3)--(-2.2,-4.3), linewidth(0.4) + wrwrwr); draw((-2.2,-4.3)--(-6.9,3.9), linewidth(0.4) + wrwrwr); draw((-8.2,-0.2)--(-5.8,1.9), linewidth(0.4) + wrwrwr); draw((-8.2,-0.2)--(-2.2,-4.3), linewidth(0.4) + wrwrwr); draw((-9.4,-4.3)--(-5.8,1.9), linewidth(0.4) + wrwrwr); draw((-6.9,-2.8)--(-8,2), linewidth(0.4) + wrwrwr); draw((-8.2,-0.2)--(-8,2), linewidth(0.4) + wrwrwr); draw((-8,2)--(-6.9,3.9), linewidth(0.4) + wrwrwr); draw((-8,2)--(-5.8,1.9), linewidth(0.4) + wrwrwr); draw(circle((-6.9,0.8), 1.6), linewidth(0.4) + wrwrwr); draw(circle((-8.9,-2.2), 2.1), linewidth(0.4) + wrwrwr); draw((-9.4,-4.3)--(-6.9,-2.8), linewidth(0.4) + wrwrwr); draw((-6.9,-2.8)--(-2.2,-4.3), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-6.9,3.9),linewidth(5pt) + dotstyle); label("$A$", (-6.9,4), NE * labelscalefactor); dot((-9.4,-4.3),linewidth(5pt) + dotstyle); label("$B$", (-9.4,-4.1), NE * labelscalefactor); dot((-2.2,-4.3),linewidth(5pt) + dotstyle); label("$C$", (-2.2,-4.1), NE * labelscalefactor); dot((-5.8,1.9),linewidth(4pt) + dotstyle); label("$Q$", (-5.8,2), NE * labelscalefactor); dot((-8.2,-0.2),linewidth(4pt) + dotstyle); label("$P$", (-8.1,-0.1), NE * labelscalefactor); dot((-8,2),linewidth(4pt) + dotstyle); label("$T$", (-8,2.1), NE * labelscalefactor); dot((-7.4,-0.8),linewidth(4pt) + dotstyle); label("$S$", (-7.3,-0.7), NE * labelscalefactor); dot((-6.9,-2.8),linewidth(4pt) + dotstyle); label("$H$", (-6.9,-2.7), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] We begin with the following claim. Claim: $H$ is the incenter of $\Delta BSC$ Proof: Note that $$BQ=BC, BH\perp QC$$$$\implies \angle SBC=\angle QBH=\angle HBC.$$Similarly, we have $$\angle SCH=\angle PCH=\angle HCB.$$So $H$ is the incenter of $\Delta BSC$. Claim: $(TPQS)$ is the cyclic Proof: Note that $$\angle PTQ=2\angle PAQ=2\angle A$$and $$\angle BSC=180-(\angle SBC+\angle SCB)=180-(\angle QBC+\angle PCB)=180-(180-2\angle C+180-2\angle B)$$$$=2(\angle C+\angle B)-180=180-2\angle A.$$ Claim: $T-S-H$ collinear Proof: As $TP=TQ\implies TS$ is the angle bisector of $\angle PSQ$. But $SH$ is the angle bisector of $\angle PSQ$. So $T-S-H$ collinear. And we are done!

Let $BE$ and $CF$ be the altitudes from $B$ and $C$ in $\Delta ABC$ $BQ=BC=CQ$ implies that $BE$ and $CF$ are the perpendicular bisectors of $CQ$ and $BP$ recpectevly. From angle chasing we get: $\angle HPS=\angle HPC=\angle HBC=\angle EBC=\angle EBQ=\angle HBS$ It follows that $BHSP$ is cyclic. $\angle ABE=\angle PBH=\angle BPH=\angle BSH=90^\circ-\angle BAC$ Analogously $\angle CSH=90^\circ-\angle BAC$ We get that: $\angle BSC=\angle PSQ=\angle BSH+\angle CSH=180^\circ-2\angle BAC$ $\angle PTQ=2\angle BAC$ Then $\angle PSQ+\angle PTQ=180^\circ$ and $PTQS$ is cyclic. $PT=QT \implies \angle TPQ=\angle TQP=90^\circ-\angle BAC$ Since $PTQS$ is cyclic, $\angle TPQ=\angle TQP=\angle TSP=\angle TSQ=90^\circ \angle BAC$ Hence $T$, $S$ and $H$ are colinear.

H is the Miquel point and the following is easy

Claim: $T$ lies on the angle bisector of $\angle PSQ$. Proof: Angle chasing gives $\angle PTQ + \angle QSP = 180^{\circ}$, so $TPSQ$ is cyclic. Since $TP = TQ$, the claim follows. Claim: $H$ lies on the angle bisector of $\angle BSC$. Proof: In fact, $H$ is the incenter of $\triangle BSC$ since the $B$ altitude and $C$ altitude are bisectors of $\angle SBC$ and $\angle SCB$, respectively. From these two claims, it follows that $T$, $H$ and $S$ are collinear.

Solution from Twitch Solves ISL: We start by eliminating $H$: Claim: The point $H$ is the incenter of $\triangle SBC$. Proof. Note that because $\overline{BH}$ is the angle bisector of $\angle SBC$ in isosceles triangle $BQC$; similarly, $\overline{CH}$ bisects $\angle SCB$. $\blacksquare$ [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (0.,4.); pair B = (-1.,-2.); pair C = (3.5,-2.); pair H = (0.,-1.41666); pair P = (-0.75675,-0.54054); pair Q = (1.21502,1.91709); pair S = (-0.26923,-0.70769); pair T = (-1.25,1.875); import graph; size(8.31315cm); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen ffqqff = rgb(1.,0.,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + blue); draw(C--A, linewidth(0.6) + zzttqq); draw(B--Q, linewidth(0.6) + blue); draw(C--P, linewidth(0.6) + blue); draw(circle(T, 2.46538), linewidth(0.6) + gray); draw(S--T, linewidth(0.6) + qqwuqq); draw(circle((0.,0.87211), 1.60258), linewidth(0.6) + ffqqff); dot("$A$", A, dir((3.455, 8.237))); dot("$B$", B, dir((-21.777, -32.757))); dot("$C$", C, dir((4.436, -30.285))); dot("$H$", H, dir((3.455, 6.953))); dot("$P$", P, dir((-28.800, -32.048))); dot("$Q$", Q, dir((12.129, 10.553))); dot("$S$", S, dir((1.542, -36.755))); dot("$T$", T, dir((-14.079, 17.234))); [/asy][/asy] Next we have the following angle chasing claim. Claim: We have $TPSQ$ cyclic. Proof. Note that \begin{align*} \measuredangle PTQ &= 2 \measuredangle PAQ = 2 \measuredangle BAC \\ \measuredangle PSQ &= \measuredangle CSB = -(\measuredangle SBC + \measuredangle BCS) = \measuredangle CBQ + \measuredangle PCB \\ &= -2\measuredangle QCB - 2\measuredangle CBP = -2 \measuredangle ACB - 2 \measuredangle CBA = 2 \measuredangle BAC. \end{align*}Hence $\measuredangle PTQ = \measuredangle PSQ$ as needed. $\blacksquare$ Hence from $TP = TQ$ we see $\overline{ST}$ is a bisector of $\angle QSP$. Since $\angle QSP > 90^{\circ}$, it follows $\overline{ST}$ is an interior angle bisector of $\angle QSP$. This concludes the proof.