Hello. Suppose that $M\equiv CL\cap AB$.Observe that $\hat{BPT}=\hat{BTP}=\hat{BQR}\Rightarrow QR\parallel BT$.It follows that $BQ=BR$.Also,it is easy to observe that $\hat{CRS}=\hat{CSR}=90^{\circ}-\frac{\hat{C}}{2}$,whence it follows that $CS=CR$.We will now apply the theorem of Menelaus. $\bullet \ \frac{LM}{LC}\cdot \frac{CR}{BR}\cdot \frac{BQ}{QM}=1\Rightarrow \frac{LM}{LC}\cdot \frac{CR}{QM}=1$ $\bullet \ \frac{LM}{LC}\cdot \frac{CS}{AS}\cdot \frac{AP}{PM}=1\Rightarrow \frac{LM}{LC}\cdot \frac{CS}{PM}=1$ Combining the last two equalities gives $PM=QM$,and we are done. [asy][asy]import graph; size(8.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 = -1.5, xmax = 8.5, ymin = -1., ymax = 6.; /* image dimensions */ pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((1.2107982688729737,4.89446940780902)--(0.,0.)--(8.,0.)--cycle, linewidth(0.8) + gray); draw(arc((0.7556771323310112,3.0547108477597367),0.5994410438149264,-103.89492078403616,-51.947460392018094)--(0.7556771323310112,3.0547108477597367)--cycle, linewidth(0.8) + qqwuqq); draw(arc((3.1467930169855514,0.),0.5994410438149264,128.05253960798194,180.)--(3.1467930169855514,0.)--cycle, linewidth(0.8) + qqwuqq); draw(arc((0.36638062816885064,1.481038437439674),0.5994410438149264,-103.89492078403616,-51.94746039201808)--(0.36638062816885064,1.481038437439674)--cycle, linewidth(0.8) + qqwuqq); /* draw figures */ draw((1.2107982688729737,4.89446940780902)--(0.,0.), linewidth(0.8) + sqsqsq); draw((0.,0.)--(8.,0.), linewidth(0.8) + sqsqsq); draw((8.,0.)--(1.2107982688729737,4.89446940780902), linewidth(0.8) + sqsqsq); draw(circle((1.2107982688729737,4.89446940780902), 1.8952168235328277), linewidth(0.4) + linetype("2 2")); draw(circle((0.,0.), 3.1467930169855514), linewidth(0.4) + linetype("2 2")); draw(circle((2.3362381725802766,1.8287217622871943), 2.0003055865671224), linewidth(0.8)); draw((-0.5035805193390578,2.592435744227195)--(2.7481610557387937,3.786154258692119), linewidth(0.8)); draw((-0.5035805193390578,2.592435744227195)--(1.5256833281750026,0.), linewidth(0.8)); draw((-0.5035805193390578,2.592435744227195)--(8.,0.), linewidth(0.8)); draw((0.7556771323310112,3.0547108477597367)--(3.1467930169855514,0.), linewidth(0.8)); /* dots and labels */ dot((1.2107982688729737,4.89446940780902),linewidth(3.pt) + dotstyle); label("$A$", (1.2930683787276627,5.022857087975251), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.4,0.16738463307435383), NE * labelscalefactor); dot((8.,0.),linewidth(3.pt) + dotstyle); label("$C$", (8.086733541963495,0.12742189682002544), NE * labelscalefactor); dot((0.7556771323310112,3.0547108477597367),linewidth(3.pt) + dotstyle); label("$P$", (0.4538509173867657,3.3244407971662957), NE * labelscalefactor); dot((2.7481610557387937,3.786154258692119),linewidth(3.pt) + dotstyle); label("$S$", (3.091391510172442,3.784012264091072), NE * labelscalefactor); dot((3.1467930169855514,0.),linewidth(3.pt) + dotstyle); label("$T$", (3.2712238233169195,-0.5), NE * labelscalefactor); dot((0.36638062816885064,1.481038437439674),linewidth(3.pt) + dotstyle); label("$Q$", (-0.3,1), NE * labelscalefactor); dot((1.5256833281750026,0.),linewidth(3.pt) + dotstyle); label("$R$", (1.3,-0.4720191469949002), NE * labelscalefactor); dot((-0.5035805193390578,2.592435744227195),linewidth(3.pt) + dotstyle); label("$L$", (-0.924863483387565,2.6850370170970415), NE * labelscalefactor); dot((0.5610288802499309,2.267874642599705),linewidth(3.pt) + dotstyle); label("$M$", (0.6,1.6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Dynamic tutorial! ACGNmath wrote: A point $P$ on the side $AB$ of a triangle $ABC$ and points $S$ and $T$ on the sides $AC$ and $BC$ are such that $AP=AS$ and $BP=BT$. The circumcircle of $PST$ meets the sides $AB$ and $BC$ again at $Q$ and $R$, respectively. The lines $PS$ and $QR$ meet at $L$. Prove that the line $CL$ bisects the segment $PQ$. Proposed by A. Antropov Let $I$ be the incenter of $\triangle ABC$ and $D$ lie on $\overline{AB}$ with $\overline{ID} \perp \overline{AB}$. Note that $I$ lies on the perpendicular bisectors of $\overline{PS}$ and $\overline{PT}$ so $I$ is the center of $\odot(PST)$. Consequently, $D$ is the midpoint of $\overline{PQ}$. Now we shall rephrase the problem: Reformulation. Let $ABC$ be a triangle with incenter $I$ and $C$-intouch point $D$; point $P$ lies on $\overline{AB}$ and $Q$ is the reflection of $P$ in $D$. Point $X$ lies on $\overline{CD}$ with $\overline{PX} \perp \overline{AI}$. Point $Y$ lies on $\overline{CD}$ with $\overline{QY} \perp \overline{BI}$. Show that $X \equiv Y$. (Proof) Move point $P$ along line $\overline{AB}$; since $X \mapsto P \mapsto Q \mapsto Y$ is linear; we only need to check for two choices of $P$. Note that $P=D$ is clear. Now we consider the position of $P$ for which $X=C$. Then we would like $Y=C$ or $Q$ to be at that position. Equivalently, let $J$ be the excenter opposite $C$ and suppose $P,Q$ lie on $\overline{AB}$ with $\overline{CP} \parallel \overline{JA}$ and $\overline{CQ} \parallel \overline{JB}$. Then we require $DP=DQ$. Let $E$ be the $C$-extouch point and $E'$ be its reflection in $J$; $M$ be the midpoint of $\overline{AB}$. Since $\overline{CD}$ passes through $E'$ and $\overline{JM} \parallel \overline{DE'}$ because midline; we conclude that $$(\overline{CP}, \overline{CQ}, \overline{CD}, \overline{C\infty})=(\overline{JA}, \overline{JB}, \overline{JM}, \overline{J\infty})=(AB;M\infty)=-1$$so $DP=DQ$ and we're done!

Let $\{S, V\} \equiv \odot(PST) \cap \overline{AC}$; observe that $E \equiv \overline{PT} \cap \overline{QV} \in \overline{CL}$ via Pascal. Hence, the conclusion is equivalent to showing that $\overline{EC}$ is the $E$-symmedian in $\triangle TEV$; however, we have $\angle RTE = \angle QPT = \angle EVT$, and so $\overline{CT}$ and $\overline{CV}$ are tangent to $\odot(TEV)$ which implies the desired.

Let $AC$ meet $\odot(PST)$ for a second time at $V.$ Because $AP = AS$, we infer that $\angle QPS = \angle USP.$ Therefore, $QPSU$ is an isosceles trapezoid. In particular, $PS \parallel QU.$ Similarly $PT \parallel QR.$ Hence, if we denote $E \equiv PT \cap QU$, then $PLQE$ is a parallelogram. Thus, $LE$ bisects $\overline{PQ}.$ The desired result now follows from Pascal's theorem for $USPTRQ$, which implies that $C, L, E$ are collinear.

\(CS, CR\) are tangent to \(\odot LSR \Rightarrow LC\) is symmedian of \(\Delta LSR\) But \(\Delta LPQ \stackrel{-}\sim \Delta LSR\), hence the conclusion.

Take a point $L'$, such $LPL'Q$ is a parallelogram. Let $U$ be second intersection of $\odot(PSQ)$ with $AC$. Since, $PSUQ$, $QPTR$ are isosceles trapezium $\implies$ $L'$ $\equiv$ $PT \cap QU$, Hence Applying Pascal on $TPSUQR$ implies the result