you sure it's $BP, CQ, BC$? because it's impossible for them to intersect at one point

i think it's perpendicular bisector of $BC$

ignitedsoulv wrote: i think it's perpendicular bisector of $BC$ indeed, this is the correct statement

Indonesia RMO 2023/5 wrote: Given a triangle $ABC$ and points $D$ and $E$ on $BC$. Points $X$ and $Y$ are inside $\triangle ABC$ such that \[ \angle BXE + \angle BCA = \angle CYD + \angle CBA = 180^{\circ} \]Suppose $AD$ intersects $XE$ at point $P$ and $AE$ intersects $YD$ at point $Q$. If $X,Y,D,E$ lies on a circle, prove that $BP, CQ$ and the perpendicular bisector of $BC$ concur. Literally angle chasing on steroids. Construct point $F = AD \cap (ABC), G = AE \cap (ABC), H = XE \cap AC, I = YD \cap AB$. Claim 01. $CFDY$, $CYQG$, $AXGH$ is cyclic and therefore by symmetry, $BXPF, BXEG, AYFI$ is cyclic. Proof. The first two are immediate, as $\angle QGC = \angle QGC = \angle ABC = \angle AFC = \angle DFC$ and $\angle DYC = \angle QYC = 180^{\circ} - \angle ABC$, the desired conclusion holds. For the last condition, we have $\angle BXH \equiv \angle BXE = 180^{\circ} - \angle BCA = \angle BCH$ and hence $BXCH$ is cyclic. Therefore, \[\angle XHA \equiv \angle XHC = \angle XBC \equiv \angle XBE = \angle XGE \equiv \angle XGA\]as desired. Claim 02. $XYGF$ is cyclic. Proof. To prove this, notice that \begin{align*} \angle GXY &= \angle GXH + \angle EXY \\ &= \angle GAC + \angle CDY = \angle GFC + \angle CFY = \angle GFY \end{align*} To finish the problem, we obtain \begin{align*} \angle QCB &= \angle YCD - \angle YCQ \\ &= \angle YFD - \angle YGQ \\ &= (\angle YFD - \angle YFX) - (\angle YGQ - \angle YGX) \\ &= \angle EGX - \angle XFD \\ &= \angle XBE - \angle XBP = \angle PBC \end{align*}and therefore, $BP, CQ$ and the perpendicular bisector of $BC$ concur. [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.5, xmax = 8.4, ymin = -4.8, ymax = 5.3; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-3.359482089532293,4.6914711178015605)--(-5.38,-0.43), wrwrwr); draw((-5.38,-0.43)--(2.415898929763407,-0.41843389927077457), wrwrwr); draw((2.415898929763407,-0.41843389927077457)--(-3.359482089532293,4.6914711178015605), wrwrwr); draw(circle((-1.4841521221447262,0.992315763827028), 4.147362151949152), wrwrwr); draw(circle((-3.4818731053025975,-2.562540805075815), 2.854928369274264), rvwvcq); draw(circle((-0.0529466733700257,-1.4009162814526892), 2.657154538864673), dbwrru); draw(circle((-2.0600140707464982,1.343990770050171), 1.8297434946058322), dotted + wvvxds); draw((-3.5566867051855553,0.2914071480164555)--(-1.5900822851907361,-0.42437722699590075), wrwrwr); draw((-2.524696649923481,-0.42576383346461855)--(-0.23267839470851523,1.2501527013830718), wrwrwr); draw(circle((-3.774988143553676,-1.821710127666104), 2.124363466731077), dotted + sexdts); draw(circle((0.21210852562790322,-0.9801894282439817), 2.2742606314342457), dotted + sexdts); draw((-3.5566867051855553,0.2914071480164555)--(5.226935433732648,-2.90556496245344), wrwrwr); draw((5.226935433732648,-2.90556496245344)--(2.415898929763407,-0.41843389927077457), wrwrwr); draw((-0.6733167493640577,-3.075012452295681)--(-3.359482089532293,4.6914711178015605), wrwrwr); draw((-2.0866399859738403,-3.1110512918187294)--(-3.359482089532293,4.6914711178015605), wrwrwr); draw((-6.535260447220864,-3.3582754601321976)--(-5.38,-0.43), wrwrwr); draw((-6.535260447220864,-3.3582754601321976)--(-2.524696649923481,-0.42576383346461855), wrwrwr); draw(circle((2.1265695010385324,2.241142879683453), 6.008399997413925), dotted + rvwvcq); draw(circle((-4.518349701255899,0.4973406485430236), 4.351287686957138), dotted + wvvxds); draw(circle((-1.4381415511501308,-0.8120655641221297), 2.388699561711658), dotted + dtsfsf); /* dots and labels */ dot((-3.359482089532293,4.6914711178015605),dotstyle); label("$A$", (-3.273343410070948,4.825571996266121), NE * labelscalefactor); dot((-5.38,-0.43),dotstyle); label("$B$", (-5.9,-0.42), NE * labelscalefactor); dot((2.415898929763407,-0.41843389927077457),dotstyle); label("$C$", (2.46,-0.42), NE * labelscalefactor); dot((-2.524696649923481,-0.42576383346461855),dotstyle); label("$D$", (-3.2,-1), NE * labelscalefactor); dot((-1.5900822851907361,-0.42437722699590075),dotstyle); label("$E$", (-1.5,-1), NE * labelscalefactor); dot((-2.0866399859738403,-3.1110512918187294), dotstyle); label("$F$", (-2.0162458852097953,-2.9527189388122657), NE * labelscalefactor); dot((-0.6733167493640577,-3.075012452295681),dotstyle); label("$G$", (-0.6020111697409989,-2.913434641160354), NE * labelscalefactor); dot((-3.5566867051855553,0.2914071480164555),dotstyle); label("$X$", (-3.469764898330503,0.4846571057299507), NE * labelscalefactor); dot((-0.23267839470851523,1.2501527013830718), dotstyle); label("$Y$", (-0.15024174674402224,1.2), NE * labelscalefactor); dot((-2.5839328767334946,-0.06264574846857822),dotstyle); label("$P$", (-2.7,0.09181412921084026), NE * labelscalefactor); dot((-1.77835085935838,0.11996185972132477), dotstyle); label("$Q$", (-1.8,0), NE * labelscalefactor); dot((5.226935433732648,-2.90556496245344), dotstyle); label("$H$", (5.26,-2.9), NE * labelscalefactor); dot((-6.535260447220864,-3.3582754601321976), dotstyle); label("$I$", (-6.8,-3.28), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Thx for the solutions and the correction