My proof proceeds in six steps. The first two steps set the stage by incorporating all the angle conditions. By the fourth step we show that $Q$ lies on $\omega$ even if $M$ is arbitrary on $\overline{AC}$. Step 5 uses the midpoint condition for the first time by projecting $(AC;M\infty)$, and the sixth steps solves what is essentially a completely projective problem. [asy][asy] size(12cm); draw(unitcircle, red); pair P = dir(0); pair R = dir(180); pair K = dir(75); pair L = conj(K); pair B = dir(65); pair D = dir(-58); pair A = extension(K, D, B, L); pair C = extension(K, B, D, L); pair T = 2/(K+L); draw(A--B--C--D--cycle, orange); draw(P--B--K--A--L--D--P, orange); draw(K--T--L, red); draw(A--T, paleblue); pair E = extension(B, P, A, C); pair F = extension(D, P, A, C); draw(F--P, orange); pair M = midpoint(A--C); pair O = midpoint(P--R); pair X = -B+2*foot(O, B, M); pair Y = -D+2*foot(O, D, M); draw(B--M--D, yellow); draw(B--Y, yellow+dashed); draw(D--X, yellow+dashed); pair Q = extension(X, E, Y, F); draw(X--Q, brown+dotted); draw(Y--F, brown+dotted); pair Z = R*Q/B; draw(Z--K--X--L--cycle, paleblue); draw(Z--T, paleblue); draw(Z--B, dashed+paleblue); draw(T--R--Q, paleblue); dot("$P$", P, dir(P)); dot("$R$", R, dir(R)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$T$", T, dir(T)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Q$", Q, dir(Q)); dot("$Z$", Z, dir(Z)); /* TSQ Source: !size(12cm); unitcircle 0.1 yellow / red P = dir 0 R = dir 180 K = dir 75 L = conj K B = dir 65 D = dir -58 A = extension K D B L C = extension K B D L T = 2/(K+L) A--B--C--D--cycle 0.1 lightred / orange P--B--K--A--L--D--P orange K--T--L red A--T paleblue E = extension B P A C F = extension D P A C F--P orange M = midpoint A--C O := midpoint P--R X = -B+2*foot O B M Y = -D+2*foot O D M B--M--D yellow B--Y yellow dashed D--X yellow dashed Q = extension X E Y F X--Q brown dotted Y--F brown dotted Z = R*Q/B Z--K--X--L--cycle paleblue Z--T paleblue Z--B dashed paleblue T--R--Q paleblue */ [/asy][/asy] Step 1: Observe that $K = \overline{BC} \cap \overline{AD}$ lies on $\omega$ by all the given angle conditions. Similarly, $L = \overline{BA} \cap \overline{CD}$ lies on $\omega$ too, and actually $P$ is the arc midpoint of $\widehat{KL}$. Step 2: Let $T = \overline{KK} \cap \overline{LL}$, and $S = \overline{XX} \cap \overline{YY}$ (not pictured). Then $\overline{ACST}$ are collinear by Brokard Theorem on $LKBD$ and $BDYX$. Step 3: Again by Brokard Theorem on $DBXY$, we find that $\overline{XD} \cap \overline{BY}$, $S$ and $M$ are collinear; in other words, $\overline{XD} \cap \overline{BY}$ lies on $\overline{AC}$. Step 4: By the converse of Pascal theorem on $BPDXQY$, we find that $Q$ lies on $\omega$ (since we already know $\overline{XD} \cap \overline{BY}$, $E$, $F$ collinear). Step 5: Let $Z$ be the point on $\omega$ such that $\overline{ZB} \parallel \overline{AC}$. Then $-1 = (AC;M\infty) = (LK;XZ)_\omega$. Thus $T \in \overline{XZ}$. Step 6: Introduce $R$ the antipode of $P$ (so $\overline{PR} \cap \overline{XZ} = T$). This is motivated because now the problem is ``completely projective'': we just need to check $\overline{BZ}$, $\overline{QR}$, $\overline{ACT}$ are parallel. Indeed, this follows by Pascal on $ZBPRQX$. Remark: Motivational remarks: I found the steps in the order 4 3 1 2 5 6. You basically have to realize $Q$ lies on the circle; then the obvious Pascal reverse-engineers to tell you about $DBXY$, which leads naturally to discovery of $K$ and $L$. In fact, I found $K$ and $L$ because I was about to try complex bashing. Once $Q$ is on the circle, you can focus on just the upper half of the diagram (ignoring $D$, $Y$, $F$ and concentrating only $B$, $X$, $E$). Adding $Z$ eliminates the midpoint as usual with harmonic arguments. Then the addition of point $R$ does two things: it changes perpendicularity to $\overline{BZ} \parallel \overline{AC} \parallel \overline{RQ}$, and it lets one just define $T = \overline{ZX} \cap \overline{PR}$, thus we can ignore $KLT$. At that point the problem is completely projective.

Let $Y'\equiv DM\cap \odot(ABC)$ such that $Y$ lies on the arc opposite $B$; I claim that $Y'\equiv Y$. Compute $$\angle BPD=\angle BY'D\iff 2\pi-\angle B-\angle BCD=2\pi-\angle CBY'-\angle BCD-\angle CDM$$$$\iff \angle B=\angle CBY'+\angle CDM \iff \angle ADY'=\angle CAY'$$which is obvious by reflection about $M$. Analogously, $X$ lies on $\odot(ADC)$; we therefore notice that $X$ and $Y$ are the $B$ and $D$ HM points in $\triangle ABC$ and $\triangle ADC$, so $B$ and $D$ are the $X$ and $Y$ HM points in $\triangle CXA$ and $\triangle CYA$ and hence by analogy $Q \in \odot(BPD)$. Let $B'$ be the reflection of $B$ over $AC$; notice that quadrilateral $BXEB'$ is cyclic on the $B$-Apollonius circle in $AC$, so $$\angle BPQ=\pi-\angle BXE=\angle B'BE=\frac{\pi-\angle BEB'}{2}=\frac{\pi}{2}-\angle BEC$$as desired.

For anyone who has seen HM points, this is a walk in the park. cjquines0 wrote: Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$. Let $T_1=EX \cap PR$ and $T_2=FY \cap PR$ where $R$ is a point on $AC$ such that $PR \perp AC$. Claim 1: $X$ is the HM point corresponding to $B$ in triangle $ABC$. (Proof) Let $ABCB_1$ be a parallelogram; $L$ be midpoint of minor arc $AC$ in $(ADC)$. Note that $$\measuredangle DXM=\measuredangle BPL=\measuredangle DLB_1,$$so $D, X, L, B_1$ are concyclic, just as we desired. $\blacksquare$ Claim 2: In any triangle $ABC$, let $X$ be the HM point opposite $B$; $M$ the midpoint of $AC$ and $E$ the foot of $B$-internal bisector. Point $E'$ lies on ray $MB$ such that $EE' \perp AC$. Then $(XM, E'B)$ depends only on $\measuredangle ABC$. (Proof) Let $N$ be midpoint of arc $ABC$, $L$ the midpoint of minor arc $AC$ in $(ADC)$. Notice that $$(XM, E'B) \overset{E}{=} (NM, \infty L)$$where projection is preceded by reflection of the pencil in line $EE'$. $\blacksquare$ To conclude, consider $(XM, E'B) \overset{E}{=} (T_1R, \infty P)$ and $(YM, F'D)\overset{F}{=}(T_2R, \infty P)$; hence, $\boxed{T_1=T_2},$ as desired. $\blacksquare$

Let $\omega_1$ be the circumcircle of $ABC$ and $\omega_2$ the circumcircle of $ADC$, then these two circles are symmetric w.r.t. $AC$. Also notice that $BP$ passes through $M_1$, the midpoint of arc $AC$ of $\omega_1$ not containing $B$, and $DP$ passes through $M_2$, the midpoint of arc $AC$ of $\omega_2$ not containing $D$. We first start with a preliminary observation: $X$ lies on $\omega_2$ and $Y$ lies on $\omega_1$. W.L.O.G. for this section we assume that $AB\le AC$. Indeed, let $X'$ be on $BM$ satisfying $MX'\cdot MB=MA^2=MC^2$. Then $\angle X'AC=\angle MBA$ and $\angle X'CA=\angle MBC$. Thus $\angle ADC=\angle ABC=\angle MBA+\angle MBC=\angle X'AC+\angle X'CA=\pi - \angle AX'C$, so $X'$ lie on $\omega_2$. In addition, let $BM$ intersect $\omega_1$ again at $X''$, then $X'$ and $X''$ are symmetrical w.r.t. $AC$. Combining with the fact that $M_1$ and $M_2$ are also symmetrical w.r.t. $AC$ (being the midpoint of arc) we have $X'M_2=X''M_1$. Knowing that the two circles have the same radius further allows us to assert $\angle X'BP=\angle X''BM_1=\angle X'DM_2=\angle X'DP$, showing that $D, B, P, X'$ cyclic hence $X'=X$. Similarly, $Y$ lies on $\omega_1$. Next, let $N_1$ be diametrically opposite $M_1$ w.r.t. $\omega_1$ and define similarly for $N_2$. We claim that $XE$ passes through $N_2$ by claiming that $XE$ is the internal angle bisector of $\angle AXC$. Indeed, by angle bisector theorem we have $\frac{AE}{EC}=\frac{AB}{BC}$. Invoking our $X''$ from the previous section (i.e. the other intersection of $BM$ and $\omega_1$) gives $AXCX''$ parallelogram. Now invoking a little bit more trigonometric bashing we have $1=\frac{AM}{CM}=\frac{AB}{BC}\cdot\frac{\sin\angle ABM}{\sin\angle CBM}=\frac{AB}{BC}\cdot\frac{AX''}{CX''}=\frac{AB}{BC}\cdot\frac{CX}{AX},$ so $\frac{AX}{CX}=\frac{AB}{BC}=\frac{AE}{EC}$, and the conclusion follows by the angle bisector theorem. Analogously, $YF$ passes through $N_1$. Finally, considering triangle $PEF$, and letting the perpendicular from $P$ to reach $AC$ at $P_1$ we have (considering signed length) $\frac{EP_1}{FP_1}=\frac{\cot\angle FEP}{\cot\angle EFP}$. Similarly if letting perpendcular from $Q$ to reach AC at $Q_1$ we have $\frac{EQ_1}{FQ_1}=\frac{\cot\angle FEQ}{\cot\angle EFQ}$. Now $\cot\angle FEP=\cot\angle MEM_1=\frac{MM_1}{EM}$, $\cot\angle EFP=\cot\angle MFM_2=\frac{MM_2}{FM}$. Considering $MM_2=MM_1$ we have $\frac{\cot\angle FEP}{\cot\angle EFP}=\frac{FM}{EM}$. Analogously, $\cot\angle FEQ=\cot\angle FEQ=\cot\angle MEN_2=\frac{MN_2}{EM}$, and $\cot\angle EFQ=\cot\angle N_1FM=\frac{MN_1}{FM}$. Therefore we have $\frac{\cot\angle FEQ}{\cot\angle EFQ}=\frac{FM}{EM}$ since again it is not hard to verify that $MN_2=MN_1$. (For signed convention we can say that $ME<0$ if it's nearer to $A$ than $B$, and $>0$ otherwise). Therefore, $\frac{EP_1}{FP_1}=\frac{EQ_1}{FQ_1}$, so $P_1\equiv Q_1$ and the two perpendicular lines coincide. (Sorry for the trigo-heavy proof).

I would like to present my proof. Let $AB\cap CD= \{J\}$, $BC\cap AD=\{K\}$. The condition of the problem implies that $BDKJ$ is cyclic and denote its circumcircle by $\gamma$. The exterior angle bisectors of $\angle JBK$ and $\angle LDK$ both pass through the midpoint of the arc $\overarc{JBDK}$. As their intersection is $P$, we conclude that $P$ is the midpoint of the arc $\overarc{JBDK}~~~(1)$. Keep this in mind. Denote by $AA'$ and $AA"$ the tangents from $A$ to $\gamma$. Let $M'$ the midpoint of $AA'$ and $M"$ the midpoint of $AA"$. It is well-known that $M'M"$ is the radical axis of $\gamma$ and the degenerated circle at $A$. Also, $A'A"$ is the polar of $A$ WRT $\gamma$, and as $C$ belongs to this polar (well-known), we get that $C\in A'A"$. As $M$ is the midpoint of $AC$, we find out that $M$ lies on $M'M"$, hence $M$ is on the radical axis of the degenerated circle at $A$ and $\gamma$. Hence $MA^2=MC^2=MX\cdot MB=MY\cdot MD$. Using this, we arrive at $\Delta MXC \sim \Delta MCB$ $\Delta MYC \sim \Delta MCD$ $\Delta MXA \sim \Delta MAB$ $\Delta MYA \sim \Delta MAD$. Puting al these together, we get that $\frac{AX}{XC}=\frac{AB}{BC}$, and also that $\frac{AY}{YC}=\frac{AD}{DC}$. Also we get that $m(\angle AXC)=m(\angle AYC)=180^o-m(\angle ABC)~~(2)$. Therefore $XE$ and $YF$ are the angle bisectors of $\angle AXC$ and $\angle AYC$. Here comes my favourite part of the problem: Note that the fact that $\angle ABC=\angle ADC$ is enough, that is, it is not used at all that they are less that $90^o$. By replacing $ABCD$ with $AXCY$, $E$ and $F$ remain at their places, $P$ becomes $Q$, and $M$ remains at its place. $X$ becomes $B$ and $B$ becomes $X$. $Y$ becomes $D$ and $D$ becomes $Y$. Denote $AX\cap CY=\{V\}$ and $AY\cap CX=\{U\}$. With all the observations above, doing everything we did to $ABCD$ the same for $AXCY$, we get that $U$ and $V$ lie on $\gamma$. Similar to (1), $Q$ is the midpoint of the arc $\overarc{VBDU}$. Acording to (2), we have that $m(\angle JBK)+m(\angle UXV)=180^o$, therefore $JK=UV$, or equivalently written $JV\parallel KU$. Now we have a trapezoid $JVUK$ inscribed in $\gamma$, $P$ being the midpoint of $\overarc{JK}$ and $Q$ the midpoint of $\overarc{VJU}$. Denote by $O$ the center of $\gamma$. Note that the common bisector line of the segmente $JV$ and $VK$ is, actually, the external angle bisector of $\angle POQ$. As $OP=PQ$, we have that $PQ\perp KU$. (3) The final step: Let $\{W\}=BC\cap AY$. $ABCY$ is cyclic, therefore $WY\cdot WA=WC\cdot WB$. $BYKU$ is cyclic, hence $WU\cdot WY=WB\cdot WK$. From these two relations we get $\frac{WC}{WK}=\frac{WA}{WU}$, thus $AC\parallel KU$. From (3), we get that $PQ\perp AC$, exactly what we wanted.

[asy][asy] unitsize(0.16inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -22.544024968129566, xmax = 27.465755540420506, ymin = -32.784146091366274, ymax = 28.47782984915346; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((3.4596513969356657,4.55499562893679)--(-10.608546536103033,-2.4244448307545143), linewidth(2) + wrwrwr); draw((-10.608546536103033,-2.4244448307545143)--(3.657558052913712,-16.274678150667455), linewidth(2) + wrwrwr); draw((3.4596513969356657,4.55499562893679)--(3.657558052913712,-16.274678150667455), linewidth(2) + wrwrwr); draw((3.4596513969356657,4.55499562893679)--(9.200195538289595,5.045956314566677), linewidth(2) + wrwrwr); draw((9.200195538289595,5.045956314566677)--(3.657558052913712,-16.274678150667455), linewidth(2) + wrwrwr); draw((-10.608546536103033,-2.4244448307545143)--(3.5469846385830346,-4.636827276866991), linewidth(2) + wrwrwr); draw((3.5006807694596165,0.23665453610085424)--(9.200195538289595,5.045956314566677), linewidth(2) + wrwrwr); draw((3.5006807694596165,0.23665453610085424)--(-1.3650775450177906,-3.8691160539462173), linewidth(2) + wrwrwr); draw(circle((-3.946454423989982,9.908080799937078), 14.016941882662707), linewidth(2) + wrwrwr); draw((-10.608546536103033,-2.4244448307545143)--(3.558604724924689,-5.859841260865332), linewidth(2) + wrwrwr); draw((9.200195538289595,5.045956314566677)--(3.558604724924689,-5.859841260865332), linewidth(2) + wrwrwr); draw((3.4596513969356657,4.55499562893679)--(3.2947608435600477,21.909724903598224), linewidth(2) + wrwrwr); draw((3.4596513969356657,4.55499562893679)--(9.90366577557506,7.7519661926992285), linewidth(2) + wrwrwr); draw((9.90366577557506,7.7519661926992285)--(9.200195538289595,5.045956314566677), linewidth(2) + wrwrwr); draw((3.4596513969356657,4.55499562893679)--(-16.07654901172654,2.8841598839167206), linewidth(2) + wrwrwr); draw((-16.07654901172654,2.8841598839167206)--(-10.608546536103033,-2.4244448307545143), linewidth(2) + wrwrwr); draw((-16.07654901172654,2.8841598839167206)--(22.765729294890875,10.16187829142016), linewidth(2) + wrwrwr); draw((22.765729294890875,10.16187829142016)--(-10.608546536103033,-2.4244448307545143), linewidth(2) + wrwrwr); draw((22.765729294890875,10.16187829142016)--(3.2947608435600477,21.909724903598224), linewidth(2) + wrwrwr); draw((22.765729294890875,10.16187829142016)--(3.5214927742155853,-1.9538087791627707), linewidth(2) + wrwrwr); /* dots and labels */ dot((3.4596513969356657,4.55499562893679),dotstyle); label("$A$", (3.627760164678306,4.973234998505071), NE * labelscalefactor); dot((-10.608546536103033,-2.4244448307545143),dotstyle); label("$B$", (-11.708572524610382,-3.2783780873608515), NE * labelscalefactor); dot((3.657558052913712,-16.274678150667455),dotstyle); label("$C$", (4.419581689397016,-15.905846597549612), NE * labelscalefactor); dot((9.200195538289595,5.045956314566677),dotstyle); label("$D$", (9.378884923161564,5.473332761284825), NE * labelscalefactor); dot((3.5469846385830346,-4.636827276866991),linewidth(4pt) + dotstyle); label("$E$", (4.002833518492432,-4.903695816395048), NE * labelscalefactor); dot((3.5006807694596165,0.23665453610085424),linewidth(4pt) + dotstyle); label("$F$", (2.7109141886882213,0.30565587922737786), NE * labelscalefactor); dot((-1.3650775450177906,-3.8691160539462173),linewidth(4pt) + dotstyle); label("$P$", (-1.9566653254431188,-2.9866543924059954), NE * labelscalefactor); dot((3.558604724924689,-5.859841260865332),linewidth(4pt) + dotstyle); label("$M$", (4.002833518492432,-6.445663918299286), NE * labelscalefactor); dot((-3.673206078366119,-4.106197457738476),linewidth(4pt) + dotstyle); label("$X$", (-3.581983191970996,-5.0287202570899865), NE * labelscalefactor); dot((7.617876917278405,1.98716542535431),linewidth(4pt) + dotstyle); label("$Y$", (6.795046263553144,2.0143232353915335), NE * labelscalefactor); dot((-6.265589433796959,-3.915676740913798),linewidth(4pt) + dotstyle); label("$Q$", (-6.457545571212625,-4.903695816395048), NE * labelscalefactor); dot((22.765729294890875,10.16187829142016),linewidth(4pt) + dotstyle); label("$G$", (22.21472858702275,10.891058524732147), NE * labelscalefactor); dot((3.5214927742155853,-1.9538087791627707),linewidth(4pt) + dotstyle); label("$H$", (3.6694349817687644,-1.611385544761675), NE * labelscalefactor); dot((3.2947608435600477,21.909724903598224),linewidth(4pt) + dotstyle); label("$Z$", (3.4610608963164724,22.226607814406545), NE * labelscalefactor); dot((-16.07654901172654,2.8841598839167206),linewidth(4pt) + dotstyle); label("$I$", (-15.917729050746681,3.2228928287759366), NE * labelscalefactor); dot((9.90366577557506,7.7519661926992285),linewidth(4pt) + dotstyle); label("$J$", (10.087356813699357,8.098846015878527), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $I=BC\cap \omega$ and $J=CD\cap\omega$. The first claim is that $I,A,D$ and $J,A,B$ are collinear. Note that \[\angle PDI = \pi-\angle PBI = \angle PBC = \angle PDA,\]so $I,A,D$ collinear. Incidentally, $\angle PDI=\angle PBJ$, so $P$ is the midpoint of arc $\widehat{IJ}$. Let $G=IJ\cap BD$. By Brokard on $IJDB$, we have that $AC$ is the polar of $G$, so $GZ$ and $GH$ are tangent to $\omega$, where $Z$ and $H$ are the intersections of $AC$ with $\omega$. By Brokard on $BDYX$, $G'=BD\cap XY$ is on the polar of $M$. But by La'Hire, $G$ is on the polar of $M$, and $G\in BD$, so $G=G'$, so $G,X,Y$ are collinear. Also, we see that (by Brokard on the same quadrilateral), that $BY\cap DX$ is on the polar of $G$, so $BY\cap DX\in AC$. Thus, by the converse Pascal on $BPDXQY$, we have that $Q\in\omega$. We claim that $X$ is the $B$-HM point of $ABC$. Let $X'$ be the HM point. We know that $(X'AB)$ is tangent to $AC$, and that $\angle AX'C=\pi-\angle ABC$ (its on $ARC$ where $R$ is the orthocenter of $ABC$), so $X'ADC$ is cyclic. Thus, \begin{align*} \angle BX'D&=\angle BX'A+\angle AX'D\\ &= \pi-\angle BAC + \angle ACD\\ &= \pi-\angle BAC + (\pi-\angle CAD-\angle ADC)\\&=2\pi-\angle BAD-\angle ADC\\&=\angle BPD, \end{align*}so $X'$ lies on $\omega$, so $X'=\omega\cap AM=X$, as desired. [asy][asy] unitsize(0.45inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -12.550513472304992, xmax = 11.10568598292777, ymin = -4.963818860356124, ymax = 10.858184671390404; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-3.1720410998173945,1.333574927567185)--(-1.9357055767518994,9.033559325606662), linewidth(2) + wrwrwr); draw((-1.9357055767518994,9.033559325606662)--(-11.9131501488594,0.8997729896494681), linewidth(2) + wrwrwr); draw((-11.9131501488594,0.8997729896494681)--(-3.1720410998173945,1.333574927567185), linewidth(2) + wrwrwr); draw((-6.469766309168775,1.1699161082445362)--(-1.9357055767518994,9.033559325606662), linewidth(2) + wrwrwr); draw(circle((-1.8111190256771401,4.210792562893538), 4.82437571673674), linewidth(2) + wrwrwr); draw((-6.469766309168775,1.1699161082445362)--(-6.255141867596467,6.0883606806294575), linewidth(2) + wrwrwr); draw((-6.047403865818704,1.9024399448072322)--(-6.255141867596467,6.0883606806294575), linewidth(2) + wrwrwr); draw(circle((-3.369607813653483,5.3145442113644545), 3.9858686688039966), linewidth(2) + wrwrwr); /* dots and labels */ dot((-3.1720410998173945,1.333574927567185),dotstyle); label("$A$", (-2.9958666793473454,1.01167567736174), NE * labelscalefactor); dot((-1.9357055767518994,9.033559325606662),dotstyle); label("$B$", (-1.874501380495403,9.183817307351209), NE * labelscalefactor); dot((-11.9131501488594,0.8997729896494681),dotstyle); label("$C$", (-12.120400754937124,0.42795127521963494), NE * labelscalefactor); dot((-6.4017999351230825,2.7274686455080905),linewidth(4pt) + dotstyle); label("$X$", (-6.851519967180737,2.7014042098783593), NE * labelscalefactor); dot((-6.469766309168775,1.1699161082445362),linewidth(4pt) + dotstyle); label("$E$", (-6.406046081335445,1.2881767099553685), NE * labelscalefactor); dot((-5.591445684467464,1.2135052211329385),dotstyle); label("$H$", (-5.5304594781222844,1.3649825523424877), NE * labelscalefactor); dot((-6.047403865818704,1.9024399448072322),linewidth(4pt) + dotstyle); label("$P$", (-5.991294532445,2.0255127968717117), E * labelscalefactor); dot((-6.255141867596467,6.0883606806294575),linewidth(4pt) + dotstyle); label("$Q$", (-6.190989722651511,6.203750622730989), NW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $X$ lies on the $B$-appolonius circle of ABC, so since $E$ is on it too and $EA$ passes through its center, we see $\angle BXE=\pi/2+\angle BEA$. But $\angle BXE=\angle QPE$, so $\angle QPE = \pi/2+\angle PEA$, so $PQ\perp AC$.

Here's my solution (in the order I found it): By an easy angle chase, we get that $U=AB \cap CD$ and $V=AD \cap BC$ lie on $\omega$. We forget about the condition that $M$ is the midpoint of $AC$ for some time, and define points $X,Y,Q$ similarly for any arbitrary point $M$ on $AC$. We claim that in fact $Q$ lies on $\omega$. Animate $M$ on $AC$. Then $M \rightarrow X \rightarrow XE \cap \omega$ and $M \rightarrow Y \rightarrow YF \cap \omega$ are projective maps. Thus, it suffices to show our claim for three positions of $M$. When $M=A$, we get that $X=U,Y=V$, and so we wish to show that $UE$ and $VF$ meet on $\omega$. Let $UE \cap \omega=Q'$. Then the result follows by Pascal on $UQ'VBPD$. Similarly, we can show the claim to be true for $M=C$. And, when $M=AC \cap \omega$, we have $X=Y=M$, which directly gives $Q \in \omega$. Thus, our claim is true. Return to the given problem (i.e. when $M$ is the midpoint of $AC$). Let $P'$ and $Q'$ be the antipodes of $P$ and $Q$ in $\omega$. Then we wish to show that $AC,P'Q,PQ'$ are concurrent. Applying Pascal on $XQ'PBP'Q$, we see that this is equivalent to proving that $T=XQ' \cap BP'$ lies on $AC$. Let $BP' \cap AC=T'$. Then, as $\measuredangle EBT'=90^{\circ}$, so $\odot (EBT')$ is the $B$-Apollonius circle of $\triangle ABC$. Also, $\measuredangle EXT=90^{\circ}=\measuredangle EBT$, which gives that $EBXT$ is cyclic. Thus, for proving $T=T'$, it suffices to show that $\odot (BEX)$ is the $B$-Apollonius circle of $\triangle ABC$. As $X$ lies on the $B$-median of this triangle, so our problem is reduced to showing that $X$ is the $B$-Humpty point of $\triangle ABC$. However, it is well known that $MA^2=MC^2=MB \cdot MX$, i.e. $AC$ is tangent to $\odot (ABX)$ and $\odot (BCX)$, which is how we define Humpty points in the first place . Hence, done. $\blacksquare$ EDIT: $400^{\text{th}}$ post on HSO

2016 IMO Shortlist G6 wrote: Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$. Solution with Kayak : We assume that $ABCD$ is not a parallelogram. Let $S=AD\cap BC,~R=AB\cap CD,~L=AX\cap CY,~ K=AY\cap XC$ and $\{U,V\}=AC\cap\omega$ as shown in the diagram below. [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 = -17.370032730607846, xmax = 6.139541560852985, ymin = -8.950203944896286, ymax = 7.400745648818856; /* image dimensions */ pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.8,-0.08)--(2,0.5), linewidth(0.4)); draw((-6.16,3.88)--(-5.8,-0.08), linewidth(0.4)); draw((-5.8,-0.08)--(-4.450656162658686,-5.33024171139331), linewidth(0.4)); draw((-4.450656162658686,-5.33024171139331)--(2,0.5), linewidth(0.4)); draw((2,0.5)--(-6.16,3.88), linewidth(0.4)); draw(circle((-8.764650673685598,-1.3671421988830121), 5.858046294391133), linewidth(0.8) + qqffff); draw((-6.16,3.88)--(-2.980235593158234,-0.44126501296832527), linewidth(0.4)); draw((-4.450656162658686,-5.33024171139331)--(-2.8045901413496854,0.14273560487399775), linewidth(0.4)); draw((-6.16,3.88)--(-1.9,0.21), linewidth(0.4)); draw((-1.9,0.21)--(-4.450656162658686,-5.33024171139331), linewidth(0.4)); draw((-2.9069134899946403,-1.4273208830994124)--(-3.9607235462495836,1.9853181724732336), linewidth(0.4)); draw((-2.9069134899946403,-1.4273208830994124)--(-2.8045901413496854,0.14273560487399775), linewidth(0.4)); draw((-2.9486521304059456,-2.0677614476898487)--(-2.9069134899946403,-1.4273208830994124), linewidth(0.4)); draw((-5.8,-0.08)--(-14.58832583745327,-0.7334908956054996), linewidth(0.4)); draw((-3.9607235462495836,1.9853181724732336)--(-4.450656162658686,-5.33024171139331), linewidth(0.4) + wrwrwr); draw((-2.9486521304059456,-2.0677614476898487)--(-6.16,3.88), linewidth(0.4) + wrwrwr); draw((2,0.5)--(-11.433793812647506,3.8474892637385096), linewidth(0.4)); draw((-2.9486521304059456,-2.0677614476898487)--(-11.433793812647506,3.8474892637385096), linewidth(0.4)); draw((-11.540362797522098,-6.525836664813006)--(-3.9607235462495836,1.9853181724732336), linewidth(0.4)); draw((-11.540362797522098,-6.525836664813006)--(2,0.5), linewidth(0.4)); draw((-5.8,-0.08)--(-6.896106866781368,4.184907010995224), linewidth(0.4)); draw((-6.16,3.88)--(-6.896106866781368,4.184907010995224), linewidth(0.4)); draw((-5.8,-0.08)--(-5.256493221081515,-6.058574568103326), linewidth(0.4)); draw((-5.256493221081515,-6.058574568103326)--(-4.450656162658686,-5.33024171139331), linewidth(0.4)); /* dots and labels */ dot((-5.8,-0.08),linewidth(2pt) + dotstyle); label("$A$", (-5.716930594788596,-0.0019230727138700656), NE * labelscalefactor); dot((-6.16,3.88),linewidth(2pt) + dotstyle); label("$B$", (-6.082996630468782,3.9637923138215188), NE * labelscalefactor); dot((2,0.5),linewidth(2pt) + dotstyle); label("$C$", (2.0721411644064722,0.5878499847708801), NE * labelscalefactor); dot((-4.450656162658686,-5.33024171139331),linewidth(2pt) + dotstyle); label("$D$", (-4.374688463961247,-5.248869584129923), NE * labelscalefactor); dot((-3.37856160483175,0.10005567553815194),linewidth(2pt) + dotstyle); label("$E$", (-3.7968273589029216,0.1811099451262248), NE * labelscalefactor); dot((-2.8045901413496854,0.14273560487399775),linewidth(2pt) + dotstyle); label("$F$", (-2.7273913034004105,0.22178394909069035), NE * labelscalefactor); dot((-2.980235593158234,-0.44126501296832527),linewidth(2pt) + dotstyle); label("$P$", (-2.890087319258271,-0.36798910839405985), NE * labelscalefactor); dot((-1.9,0.21),linewidth(2pt) + dotstyle); label("$M$", (-1.8122262141999455,0.2827949550373886), NE * labelscalefactor); dot((-3.9607235462495836,1.9853181724732336),linewidth(2pt) + dotstyle); label("$X$", (-3.8866004163876657,2.0724511294738717), NE * labelscalefactor); dot((-2.9486521304059456,-2.0677614476898487),linewidth(2pt) + dotstyle); label("$Y$", (-2.869750317276038,-1.994949266972681), NE * labelscalefactor); dot((-2.9069134899946403,-1.4273208830994124),linewidth(2pt) + dotstyle); label("$Q$", (-2.829076313311573,-1.3441652035412326), NE * labelscalefactor); dot((-14.58832583745327,-0.7334908956054996),linewidth(2pt) + dotstyle); label("$U$", (-14.502515451113059,-0.6527071361453185), NE * labelscalefactor); dot((-11.433793812647506,3.8474892637385096),linewidth(2pt) + dotstyle); label("$K$", (-11.350280143867012,3.9231183098570535), NE * labelscalefactor); dot((-11.540362797522098,-6.525836664813006),linewidth(2pt) + dotstyle); label("$L$", (-11.451965153778175,-6.4487527010816565), NE * labelscalefactor); dot((-5.256493221081515,-6.058574568103326),linewidth(2pt) + dotstyle); label("$R$", (-5.167831541268317,-5.981001655490303), NE * labelscalefactor); dot((-3.09873970725935,0.12086294484481758),linewidth(2pt) + dotstyle); label("$V$", (-3.312109331151666,0.20144694710845756), NE * labelscalefactor); dot((-6.896106866781368,4.184907010995224),linewidth(2pt) + dotstyle); label("$S$", (-6.8151287018291535,4.26884734355501), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider the following lemmas: Lemma 1. Let $ABCD$ be a cyclic quadrilateral whose diagonals intersect at $Y$ and $AD\cap BC=X.$ Let $P$ be a point on $XY$ and the second intersection of $\odot (ABCD)$ with $PA$ and $PB$ be $K$ and $L$ respectively. Then it follows that $XY, BK$ and $AL$ are concurrent. Proof. Let $XY$ intersect $\odot(ABCD)$ at $I$ and $J.$ By Brocard's Theorem, it is easy to see that $(I,J;A,B)=-1.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); 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 = -5.172652771186477, xmax = 10.198530855245242, ymin = -4.859595378688876, ymax = 5.831089427237665; /* image dimensions */ /* draw figures */ draw(circle((2.235675488430095,-0.6187792711653043), 3.3775654652468465), linewidth(0.4) + blue); draw((-0.08,1.84)--(5.6,-0.32), linewidth(0.4)); draw((3.32,2.58)--(-1.138688740973718,-0.7657980432597458), linewidth(0.4)); draw((1.276600275234065,5.179061053430356)--(1.5220063065799334,1.23078633411749), linewidth(0.4)); draw((1.3536390081741951,3.9396046309159387)--(-0.8973864599809147,0.6429145504281734), linewidth(0.4)); draw((1.276600275234065,5.179061053430356)--(-1.138688740973718,-0.7657980432597458), linewidth(0.4)); draw((1.276600275234065,5.179061053430356)--(5.6,-0.32), linewidth(0.4)); draw((4.845512684584302,1.5252119325475504)--(1.3536390081741951,3.9396046309159387), linewidth(0.4)); draw((1.5220063065799334,1.23078633411749)--(1.8454944779205704,-3.9737319162954723), linewidth(0.4)); /* dots and labels */ dot((-0.08,1.84),linewidth(2pt) + dotstyle); label("$A$", (-0.026763442960632506,1.8952154190856052), NE * labelscalefactor); dot((3.32,2.58),linewidth(2pt) + dotstyle); label("$B$", (3.3772356992249386,2.6265433597895353), NE * labelscalefactor); dot((5.6,-0.32),linewidth(2pt) + dotstyle); label("$C$", (5.651000751231707,-0.27217465972786026), NE * labelscalefactor); dot((-1.138688740973718,-0.7657980432597458),linewidth(2pt) + dotstyle); label("$D$", (-1.0905131748936234,-0.7109714241502183), NE * labelscalefactor); dot((1.276600275234065,5.179061053430356),linewidth(2pt) + dotstyle); label("$X$", (1.329517465253931,5.232730203025358), NE * labelscalefactor); dot((1.5220063065799334,1.23078633411749),linewidth(2pt) + dotstyle); label("$Y$", (1.568861154938854,1.2835593232241365), NE * labelscalefactor); dot((1.3536390081741951,3.9396046309159387),linewidth(2pt) + dotstyle); label("$P$", (1.4092986951489053,3.9961211396532588), NE * labelscalefactor); dot((-0.8973864599809147,0.6429145504281734),linewidth(2pt) + dotstyle); label("$K$", (-0.8378726135595381,0.6984969706609924), NE * labelscalefactor); dot((4.845512684584302,1.5252119325475504),linewidth(2pt) + dotstyle); label("$L$", (4.893079067229451,1.5760904995057083), NE * labelscalefactor); dot((1.4330467410096437,2.6620339764674266),linewidth(2pt) + dotstyle); label("$I$", (1.4890799250438795,2.7196214613336718), NE * labelscalefactor); dot((1.8454944779205704,-3.9737319162954723),linewidth(2pt) + dotstyle); label("$J$", (1.9012829461679137,-3.915517491598348), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now inverting around $P$ with radius $\sqrt{PI\cdot PJ},$ we obtain that $(I,J;K,L)=(J,I;A,B)=-1,$ which implies that $AB, CD, KL$ are concurrent, thus again by Brocard, $XY, BK$ and $AL$ are concurrent. $\square$ Lemma 2. Let $ABCD$ be a quadrilateral inscribed in the circle $\Gamma, $ Let $AC\cap BD= Y$ and $AD\cap BC=X.$ Let $P$ be a point on line $XY,$ let $K$ be the second intersection of $PA$ with $\Gamma$ and $L$ be the second intersection of $XK$ with $\Gamma$, finally let $DL\cap XY= Z$ then it follows that $(X,Y;P,Z)=-1.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -22.680500494225463, xmax = 17.0747406351982, ymin = -14.444614950563711, ymax = 13.205224035644418; /* image dimensions */ /* draw figures */ draw(circle((-3.64,-1.16), 5.381152292957336), linewidth(0.4) + blue); draw((-3.722093369329189,11.091310229368936)--(-8.790835128644087,-2.7174650806763028), linewidth(0.4)); draw((-8.790835128644087,-2.7174650806763028)--(-0.4303602250270546,3.1591448823710016), linewidth(0.4)); draw((-6.56,3.36)--(1.6846835554896908,-1.9375249410132471), linewidth(0.4)); draw((1.6846835554896908,-1.9375249410132471)--(-3.722093369329189,11.091310229368936), linewidth(0.4)); draw((-3.722093369329189,11.091310229368936)--(-3.2083861761728762,1.2064597573855096), linewidth(0.4)); draw((-3.722093369329189,11.091310229368936)--(-5.1074720182720705,4.017192856711877), linewidth(0.4)); draw((-6.951948556963188,-5.401202277188562)--(-5.1074720182720705,4.017192856711877), linewidth(0.4)); draw((-6.56,3.36)--(-3.3947298101688306,4.79211902592437), linewidth(0.4)); draw((-8.790835128644087,-2.7174650806763028)--(-2.5295738376256147,-11.855374989077417), linewidth(0.4)); draw((-2.5295738376256147,-11.855374989077417)--(-3.2083861761728762,1.2064597573855096), linewidth(0.4)); /* dots and labels */ dot((-6.56,3.36),linewidth(2pt) + dotstyle); label("$A$", (-6.413866364279621,3.50714618227291), NE * labelscalefactor); dot((-0.4303602250270546,3.1591448823710016),linewidth(2pt) + dotstyle); label("$B$", (-0.2923845986763235,3.3008041002862822), NE * labelscalefactor); dot((1.6846835554896908,-1.9375249410132471),linewidth(2pt) + dotstyle); label("$C$", (1.8054265681877277,-1.788967255383871), NE * labelscalefactor); dot((-8.790835128644087,-2.7174650806763028),linewidth(2pt) + dotstyle); label("$D$", (-8.649238919134758,-2.5799452363326107), NE * labelscalefactor); dot((-3.722093369329189,11.091310229368936),linewidth(2pt) + dotstyle); label("$X$", (-3.5938579104623716,11.244974256771453), NE * labelscalefactor); dot((-3.3947298101688306,4.79211902592437),linewidth(2pt) + dotstyle); label("$P$", (-3.2499544404846583,4.9171504091815335), NE * labelscalefactor); dot((-5.1074720182720705,4.017192856711877),linewidth(2pt) + dotstyle); label("$K$", (-4.969471790373225,4.360562775230565), NE * labelscalefactor); dot((-3.2083861761728762,1.2064597573855096),linewidth(2pt) + dotstyle); label("$Y$", (-3.0780027054958015,1.340554321413318), NE * labelscalefactor); dot((-6.951948556963188,-5.401202277188562),linewidth(2pt) + dotstyle); label("$L$", (-6.826550528252877,-5.262392302158773), NE * labelscalefactor); dot((-2.5295738376256147,-11.855374989077417),linewidth(2pt) + dotstyle); label("$Z$", (-2.390195765540375,-11.727777537739778), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof. Note that the lemma is purely projective, so let us take a projective transformation fixing $\Gamma$ and taking $Y$ to the center of the circle. Now, the problem is reduced to the following : "Let $ABCD$ be a rectangle, $P$ be a point on the perpendicular bisector of $AB,$ let $K$ be the second intersection of $PA$ with $\Gamma.$ Let $L$ be the point on $\Gamma$ such that $AKLD$ is a trapezium and $Z$ be the point at which the perpendicular bisector of $AB$ intersects $LD.$ Prove that $PY=ZY.$" But now this is straightforward symmetry. Hence the lemma. $\square$ Now consider the following claims: Claim 1. $S$ and $R$ both lie on $\omega.$ Proof. Angle chasing. $\square$ Claim 2. $Q$ lies on $\omega.$ Proof. Using Lemma 1 on cyclic quadrilateral $SBDR,$ we get $BY$ and $DX$ intersect on $AC.$ Now, letting $XE\cap \omega=Q',$ and applying Pascal's Theorem on cyclic hexagon $PBYQ'XD$ we get $Q',Y,F$ are collinear. Thus $Q=Q'.$ $\square$ Claim 3. $XYLK$ is cyclic. Proof. Applying Brocard's theorem on quadrilateral $SBDR,$ we obtain $(B,D; U,V)=-1.$ Therefore, $(U,V;A,C)\overset{S}{=}(U,V;D,B)=-1,$ now since $M$ is the midpoint of $AC,$ therefore, $MA^2=MV\cdot MU=MX\cdot MB.$ Hence, $X$ is the $B-$Humpty point of $\triangle ABC$ and $Y$ is the $D-$Humpty point of $\triangle ADC,$ thus, \[\angle AXC=180^{\circ}-\angle ABC=180^{\circ}-ADC=\angle AYC\]So, the claim follows. $\square$ Claim 4. $K$ and $L$ lie on $\omega.$ Proof. Suppose $K$ and $L$ both doesn't lie on $\omega.$ Let the circumcircle of $XYLK$ be $\omega'.$ Let $\omega'$ intersect $AC$ at $U'$ and $V',$ again by Brocard, we have $(U',V';A,C)=-1$ also since $AM=MC$ so $MA^2=MV'\cdot MU'.$ But we also have $MA^2=MV\cdot MU,$ therefore, $M$ lies on the radical axis of $\omega'$ and $\omega$ which means $X,Y,M$ are collinear, which is a contradiction to the assumption that $ABCD$ is not a parallelogram. Thus, the claim follows. $\square$ Claim 5. $P$ is the mid-arc point of $SR$ in $\omega$ and $Q$ is the mid-arc point of $KL$ in $\omega.$ Proof. Note that $BP$ is the exterior angle bisector of $\angle SBR$ of triangle $SBR, $ therefore the first part follows. Similarly, as the $A-$Humpty point lies on the $A-$apollonian circle, we similarly have that $XQ$ is the exterior angle bisector of $\angle KXL$ of triangle $KXL.$ $\square$ Coming back to the original problem, Applying Lemma 2 on quadrilateral $KXYL,$ we obtain $(A,C;M,KB\cap AC)=-1$ and $(A,C;M, RL\cap AC)=-1$ thus, it follows that $KB\parallel AC\parallel LR.$ So, it suffices to solve the following problem: "$KBRL$ is a cyclic trapezium. Let $P$ and $Q$ be the mid-arc points of minor arc $SR$ and major arc $KL$ respectively. Prove that $PQ$ is perpendicular to the parallel sides." Let the center of $\omega$ be $O.$ Let the common perpendicular bisector of the parallel sides of the trapezium be $\ell$. Let $P'$ and $Q'$ be the antipodes of $P$ and $Q$ with respect to $\omega$ respectively. Note that by symmetry, $P$ and $Q'$ are reflections of each other about $\ell$. Let the midpoint of $PQ'$ be $N,$ it is obvious that $N\in\ell.$ also midpoint of $QQ'$ is $O,$ so $PQ\parallel NO$ but $NO$ is same as $\ell.$ Thus, we are done. $\blacksquare$

cjquines0 wrote: Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$. Let $\gamma_1,\gamma_2$ be the circles $(ABC),(ADC)$ respectively. Since $\angle ABC=\angle ADC,$ hence $\gamma_1,\gamma_2$ are symmetric about $AC.$ Redefine $X,Y$ so that the rays $MB,MD$ meet $\gamma_1,\gamma_2$ again in $Y,X$ respectively. Claim 1: $X,B,D,Y$ are concyclic, say lie on $\omega.$ Further, $P \in \omega.$ Proof: The key observation is that if $MD \cap \gamma_1=\{Y,Y'\},$ then $Y'$ is the reflection of $D$ over $M.$ This is because the orthocenter of $YAC$ is concyclic with $A,D,C.$ Similarly define $X'.$ Then we have $$MX \cdot MB=MX \cdot MX'=MY \cdot MY'=MY \cdot MD$$Hence $X,B,D,Y$ are concyclic. [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 = -12.727459780479958, xmax = 21.114893618568733, ymin = -9.175235342663875, ymax = 12.688663761003287; /* image dimensions */ pen qqzzff = rgb(0,0.6,1); pen ffwwzz = rgb(1,0.4,0.6); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen ttffqq = rgb(0.2,1,0); pen wwqqcc = rgb(0.4,0,0.8); draw((9.046453683703144,1.2536626656814498)--(-0.9113304425574502,-0.6856077812488472)--(-2.39839621753842,1.4035333802046477)--(2.1163306099460035,4.904798168816871)--cycle, linewidth(0.4) + ffwwzz); /* draw figures */ draw(circle((3.29,-1.27), 6.285350607784254), linewidth(0.4) + qqzzff); draw((-2.39839621753842,1.4035333802046477)--(9.046453683703144,1.2536626656814498), linewidth(0.4)); draw((9.046453683703144,1.2536626656814498)--(-0.9113304425574502,-0.6856077812488472), linewidth(0.4) + ffwwzz); draw((-0.9113304425574502,-0.6856077812488472)--(-2.39839621753842,1.4035333802046477), linewidth(0.4) + ffwwzz); draw((-2.39839621753842,1.4035333802046477)--(2.1163306099460035,4.904798168816871), linewidth(0.4) + ffwwzz); draw((2.1163306099460035,4.904798168816871)--(9.046453683703144,1.2536626656814498), linewidth(0.4) + ffwwzz); draw(circle((-4.937316151445251,5.109857787050801), 7.056626820338141), linewidth(0.4) + red); draw(circle((3.3580574661647185,3.9271960458860957), 6.285350607784249), linewidth(0.4) + qqzzff); draw((0.5478525794644877,9.549329160420113)--(3.3240287330823617,1.3285980229430487), linewidth(0.4) + linetype("2 2") + yqqqyq); draw((3.3240287330823617,1.3285980229430487)--(-2.9825697890902756,-1.6706251412055315), linewidth(0.4) + linetype("2 2") + yqqqyq); draw((0.5478525794644877,9.549329160420113)--(2.4285521212741736,1.3403243208681919), linewidth(0.4) + linetype("2 2") + ttffqq); draw((-2.9825697890902756,-1.6706251412055315)--(1.93349868099754,3.5011671331960197), linewidth(0.4) + linetype("2 2") + ttffqq); draw((1.9732671696014106,6.538080544321976)--(-0.9113304425574502,-0.6856077812488472), linewidth(0.4)); draw((3.4403572987472604,10.212007817159023)--(1.9732671696014106,6.538080544321976), linewidth(0.4)); draw((3.4403572987472604,10.212007817159023)--(3.2077001674174603,-7.55481177127293), linewidth(0.4) + wwqqcc); draw((1.9732671696014106,6.538080544321976)--(3.2077001674174603,-7.55481177127293), linewidth(0.4)); draw((-3.5207022920809425,12.022830176332599)--(3.3240287330823617,1.3285980229430487), linewidth(0.4) + blue); draw((2.4285521212741736,1.3403243208681919)--(3.2757576335821823,-2.357615725386829), linewidth(0.4) + linetype("2 2") + ttffqq); draw((1.93349868099754,3.5011671331960197)--(3.372299832582537,5.01481177127293), linewidth(0.4) + linetype("2 2") + ttffqq); draw((1.7965827708801083,7.219499317126694)--(3.3240287330823617,1.3285980229430487), linewidth(0.4) + blue); draw((3.3240287330823617,1.3285980229430487)--(4.531726856218719,-2.2476021229307683), linewidth(0.4) + linetype("2 2") + yqqqyq); draw((3.3240287330823617,1.3285980229430487)--(7.559387908722176,3.342803827134945), linewidth(0.4) + linetype("2 2") + yqqqyq); /* dots and labels */ dot((-2.39839621753842,1.4035333802046477),linewidth(4pt) + dotstyle); label("$A$", (-3.2284167236584875,1.0966112170854985), NE * labelscalefactor); dot((9.046453683703144,1.2536626656814498),linewidth(4pt) + dotstyle); label("$C$", (9.3940404908297,0.8390100494428809), NE * labelscalefactor); dot((2.1163306099460035,4.904798168816871),linewidth(4pt) + dotstyle); label("$B$", (2.3100083806577585,5.186029753412052), NE * labelscalefactor); dot((-0.9113304425574502,-0.6856077812488472),linewidth(4pt) + dotstyle); label("$D$", (-1.0388067986962506,-1.7692017729386216), NE * labelscalefactor); dot((1.9732671696014106,6.538080544321976),linewidth(4pt) + dotstyle); label("$P$", (2.2456080887471046,6.152034132071868), NE * labelscalefactor); dot((2.4285521212741736,1.3403243208681919),linewidth(4pt) + dotstyle); label("$E$", (1.5694050236852373,0.8068099034875538), NE * labelscalefactor); dot((-0.08916004100671432,1.3732938507423975),linewidth(4pt) + dotstyle); label("$F$", (-0.45920417150036447,1.6762138442813879), NE * labelscalefactor); dot((3.3240287330823617,1.3285980229430487),linewidth(4pt) + dotstyle); label("$M$", (3.662414510781493,0.7746097575322266), NE * labelscalefactor); dot((-2.9825697890902756,-1.6706251412055315),linewidth(4pt) + dotstyle); label("$Y$", (-3.5182180372564305,-2.4454048380004925), NE * labelscalefactor); dot((0.5478525794644877,9.549329160420113),linewidth(4pt) + dotstyle); label("$X$", (0.37799962333813786,10.080451938621785), NE * labelscalefactor); dot((1.93349868099754,3.5011671331960197),linewidth(4pt) + dotstyle); label("$Q$", (1.1186029803106592,3.2862211420477476), NE * labelscalefactor); dot((3.4403572987472604,10.212007817159023),linewidth(4pt) + dotstyle); label("$T$", (3.340413051228223,10.692254711773002), NE * labelscalefactor); dot((3.2077001674174603,-7.55481177127293),linewidth(4pt) + dotstyle); label("$S$", (2.9862114457196256,-8.499032277602005), NE * labelscalefactor); dot((-3.5207022920809425,12.022830176332599),linewidth(4pt) + dotstyle); dot((3.372299832582537,5.01481177127293),linewidth(4pt) + dotstyle); label("$K$", (3.501413781004858,5.282630191278034), NE * labelscalefactor); dot((3.2757576335821823,-2.357615725386829),linewidth(4pt) + dotstyle); label("$L$", (3.598014218870839,-3.153808049017691), NE * labelscalefactor); dot((1.7965827708801083,7.219499317126694),linewidth(4pt) + dotstyle); dot((7.559387908722176,3.342803827134945),linewidth(4pt) + dotstyle); label("$Y'$", (7.68743275519737,3.6082226016010193), NE * labelscalefactor); dot((4.531726856218719,-2.2476021229307683),linewidth(4pt) + dotstyle); label("$X'$", (4.66061903539663,-3.1216079030623636), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We show $P \in (XBD).$ It is easy to see that $\measuredangle MBE=\measuredangle ELM,$ since they subtend equal arcs in their circles. Hence $$\measuredangle XBP=\measuredangle MBE=\measuredangle ELM=\measuredangle XLT=\measuredangle XDP$$Hence we are done here. $\square$ Let the perpendicular bisector of $AC$ meet $\gamma_1,\gamma_2$ in $\{S,K\},\{T,L\}$ respectively, so that $K,L$ lie in the interiors of $\gamma_2,\gamma_1$ respectively. Claim 2: The point $Q$ lies on $\omega.$ Further, $K,L$ lie on $YQ, XQ$ respectively. Proof: Firstly, see that $D$ is the $Y$-HM point (The Humpty point) in $\triangle YAC.$ This is because it lies on the $Y$ median and $\measuredangle ADC=-\measuredangle AYC.$ Hence, by a well-known lemma, it lies on the $Y$ Apollonius circle and so $YA:YC=DA:DC=AF:FC,$ so $YF$ bisects $\angle AYC.$ Since $K$ is the arc midpoint in $\gamma_1,$ hence $YF$ passes through $K.$ Due to the angle bisectors, $P=DT \cap BS,$ as $T,S$ are arc midpoints. Now notice that $EMXT, EMSY$ are both cyclic, due to right angles. Hence $$\measuredangle BXQ+\measuredangle BYQ = \measuredangle MXE+\measuredangle BYM=\measuredangle MTE+\measuredangle BSM = \measuredangle BPD$$Hence $Q \in \omega,$ which is what we wanted. $\square$ To conclude, notice that $M$ is the midpoint of both $KL, TS.$ Let $PQ \cap TS=Z.$ Our goal is to show $Z=\infty.$ Then $$(T,S;M,Z) \overset{P}{=} (D,B; PM \cap \omega, Q) \overset{M}{=} (Y,X;P,MQ \cap \omega) \overset{Q}{=} (K,L;M,Z)$$Thus $ZT:ZS=ZL:ZK.$ It is not too hard to check now, say algebraically that since $\{T,S\},\{K,L\}$ have the same midpoint, this is possible if and only if $Z \in \{M,\infty\}.$ Now, $P,M,Q$ are collinear happens only if $P, Q \in TS$ (Proved formally in the comments). Even here, $PQ, TS$ are "parallel", since they coincide. Hence we are done. $\blacksquare$ Comments: We present a formal way to see that $P,M,Q$ are collinear happens only twice, which is when $P,Q \in TS$ (note that $P$ lands on $TS$ exactly twice, when $B=K,S).$ Move $B$ on its circle $\gamma_1$ while fixing the other points (in particular $D).$ Then $B \overset{M}{\mapsto} P$ is projective and so $P$ moves oN $TD$ with degree $1.$ Also, $B \overset{M}{\mapsto} X \overset{L}{\mapsto} Q$ is projective and so $Q$ moves on $YK$ with degree $1$ (note that $Y$ is fixed as $D$ is fixed). Hence, by Zack's lemma, $\deg MP \le 1$ and $\deg MQ \le 1.$ Now assume on the contrary that $M,P,Q$ are collinear for more than three positions of $B.$ Then by Zack's lemma $0=\deg M=\deg (MP \cap MQ) \le 1+1-3=-1,$ a contradiction. Hence, for every position of $D,$ there are exactly two positions of $D$ for which $M,P,Q$ are collinear. And even in these two positions, $PQ \parallel TS,$ and hence we are home free!

We will deal with non-trivial cases where $(AB, CD)$ and $(AD, BC)$ aren't parallel. Denote $\Omega$ by the circumcircle of triangle $BPD$, and $R$ by the antipodal point of $P$ on $\Omega$. Let $U=AB \cap CD$, and $V=AD \cap BC$, $E' = AC \cap BR$, $F' = AC \cap DR$. Since $\angle PBV=\angle PDA$, $V$ lies on $\Omega$. Similarly $U$ lies on $\Omega$ too. Note that $\angle PUV=\angle PBV=\angle PVU$, so $P$ is the midpoint of arc $UV$. Note that $\odot (AC)$ and $\Omega$ are orthogonal, so $MC^2=MY \cdot MD$. Since $DF \perp DF'$, $D(AC, FF')=-1$, so $MC^2=MF \cdot MF'$. Combining these results give $MY \cdot MD=MF \cdot MF'$, so $(YDFF')$ are concyclic. Similarly $(XBEE')$ are concyclic too. Hence, $\angle QFE=\angle YFF'=\angle RXY$, and similarly we get $\angle QEF=\angle RYX$. This leads to $\angle XQY=\angle XRY$, thus $Q$ lies on $\Omega$. Then $\angle QEF=\angle RYX=\angle RQX$, so $RQ$ is parallel to $AC$, but since $PQ\perp QR$, we obtain $PQ \perp AC$, and we're done.

By an angle chase, notice that $G $, $H $ lies on $\odot (PBD)$. $BI $ is the external angle bisector of $\angle ABC $. By Braikenridge-Maclaurin's theorem(converse of Pascal's theorem) $ME.MI=MX.MB $, hence $IBXE $ is concyclic quadrilateral, and $\angle IXE=90^{\circ} $. Let $Q_1$ be the second intersection of $XE $ with $\odot (PBD) $. Now angle chasing $PQ_1\perp CA $.

A wonderful problem rich in configuration. Here is my solution with some remarks on motivation. Use directed angles modulo $180^{\circ}$ to avoid configuration issues. Let $T=AB\cap CD$ and $S=AD\cap BC$. Since $\measuredangle SBT=\measuredangle CBA=\measuredangle ADC=\measuredangle SDT$, we obtain that $SDBT$ is cyclic. Furthermore, since $BP$ is the external angle bisector of $\angle SBT$ and $DP$ is the external angle bisector of $\angle SDT$, this means that $P$ is the midpoint of arc $\overarc{SDBT}$.

We claim that $A, B, C, Y$ are concyclic. Define $Y_1$ to be the second intersection point of $(ABC)$ and $(SDBT)$. It now suffices to show that $D, Y_1, M$ are collinear. By Pascal's Theorem on the hexagon $SSDTTB$, we get that $SS\cap TT$ lies on $AC$. Now let $W=SS\cap TT$. Also let $WY_1\cap (SDBT)=V'$. Since $\measuredangle WCY_1=\measuredangle ACY_1=\measuredangle ABY_1=\measuredangle TBY_1=\measuredangle WTY_1$, we get $WTCY_1$ is cyclic. This implies that $\measuredangle DVY_1=\measuredangle DTY_1=\measuredangle CWY'$, or that $DV\parallel AC$. Now $-1=(Y_1,V;S,T)\stackrel{D}{=}(DY'\cap AC,P_\infty;A,C)$, implying that $DY_1\cap AC$ is the midpoint of $AC$. Similarly, $ADCX$ is cyclic.

Now let $Y'=AY\cap CX$ and $X'=AX\cap YC$. We claim that $X'$ and $Y'$ are on $(SDBT)$. By radical axis theorem on $(SDBT)$, $(ABC)$ and $(ADC)$, we note that $DX \cap BY$ is on $AC$. By the converse of Pascal's Theorem on $DSBYY'X$, since $A, C$ and $DX\cap BY$ are collinear, $Y'$ lies on $(DSBYX)$. Similarly, $X'$ lies on the circle.

We claim that $EX$ bisects $\angle AXC$. To see why this is true, note that $\measuredangle ACX=\measuredangle ADX=\measuredangle SDX=\measuredangle SBX$. Then, $\measuredangle MCX=\measuredangle CBM$ and $MX\cdot MB=MC^2$. Now let $H$ be the midpoint of arc $\overarc{AC}$ on $(ADC)$ not containing $X$ and $I$ be the midpoint of $\overarc{AC}$ on $(ABC)$ not containing $B$. Since the circles $(ABC)$ and $(ADC)$ are symmetric with respect to $AC$, letting $I'$ be the reflection of $I$ about $AC$, we observe that $I'$ lies on $(ACD)$. Since $M=BX\cap HI$, we get $$MC^2=MA\cdot MC=MI'\cdot MH=MI\cdot MH$$which means that $HIXB$ is cyclic. Thus, by radical axis theorem on $(HIXB)$, $(ABC)$, $(ADC)$, we know $BI\cap XH$ is on $AC$. We can make our next claim that $Q$ is the arc midpoint of $\overarc{X'TSY'}$. Then since $QY$ is the external angle bisector of $\angle Y'YX'$, it bisects $\angle AYC$. Similarly, $QX$ bisects $\angle AXC$. Hence $Q$ is the arc midpoint. This means that $PQ$ is the external angle bisector of $\angle X'PY'$.

[asy][asy] size(10cm); pair B=dir(60); pair T=dir(40); pair D=dir(160); pair S=dir(190); draw(B--T--D--S--cycle); pair A=extension(B,T,D,S); pair C=extension(B,S,T,D); draw(A--D); draw(A--B); draw(A--C); pair O1=circumcenter(S,D,B); filldraw(CR(O1, abs(D-O1)), invisible, lightred); filldraw(CR(circumcenter(A,B,C), abs(A-circumcenter(A,B,C))), invisible, lightred); pair E1=bisectorpoint(A,B,C); pair E=extension(A,C,B,E1); pair F1=bisectorpoint(A,D,C); pair F=extension(A,C,D,F1); pair P=extension(D,F,B,E); pair M=(A+C)/2; pair O = circumcenter(D,B,T); pair Y = -D+2*foot(O, D, M); pair X = -B+2*foot(O,B,M); pair Xp=extension(A,X,Y,C); pair Yp=extension(A,Y,X,C); draw(D--M); draw(B--M); pair Q=extension(X,E,Y,F); draw(A--Xp); draw(A--Yp); filldraw(CR(circumcenter(A,D,C), abs(A-circumcenter(A,D,C))), invisible, lightred); path p1 = circle(circumcenter(A,D,C), abs(A-circumcenter(A,D,C))); path q1=(11*M-10*circumcenter(A,D,C))--(-10*M+11*circumcenter(A,D,C)); path p2 = circle(circumcenter(A,B,C), abs(A-circumcenter(A,B,C))); path q2 = (11*M-10*circumcenter(A,B,C))--(-10*M+11*circumcenter(A,B,C)); pair Ip=intersectionpoints(p1,q1)[0]; pair H=intersectionpoints(p1,q1)[1]; pair I=intersectionpoints(p2,q2)[1]; draw(H--Ip); draw(C--Xp); draw(C--Yp); dot("$A$",A,dir(A));dot("$B$",B,dir(B));dot("$C$",C,dir(C));dot("$D$",D,dir(D));dot("$E$",E,dir(E));dot("$F$",F,dir(F));dot("$S$",S,dir(S));dot("$T$",T,dir(T));dot("$P$",P,dir(P));dot("$Y$",Y,dir(Y));dot("$X$",X,dir(X));dot("$Y'$",Yp,dir(Yp));dot("$X'$",Xp,dir(Xp));dot("$Q$",Q,dir(Q));dot("$I'$",Ip,dir(Ip));dot("$H$",H,dir(H));dot("$I$",I,dir(I));dot("$M$",M,dir(M)); [/asy][/asy] The finish line is now in sight. Note that since $\measuredangle X'TD=\measuredangle X'XD=\measuredangle AXD=\measuredangle ACD=\measuredangle ACT$, we get the $X'T \parallel AC$. Similarly, $Y'S \parallel AC$. Thus $Y'S \parallel X'T$. This also means that the two lines are parallel to the line segment connecting the arc midpoints of $\overarc{ST}$ and $\overarc{X'Y'}$. Let $Q'$ be the midpoint of $\overarc{X'Y'}$, then since $QQ'$ is a diameter of the circle, it follows that $PQ' \perp PQ$, and since $PQ'$ is parallel to $AC$, the conclusion follows.

The main strategy in solving this problem is to make lots of observations. There is no observation which is hard to prove and at the end we finish it by very little projective geometry. The one thing I am using is the HM point , so I would like to start by writing some of its properties , let $T_A$ be the $A-HM point$ - 1. $T_A\in\odot(BHC)$ where $H$ is orthocenter 2. $T_A$ lies on the $A-median$ 3. $T_A$ lies on $A-$apollonious circle. 4. $\sphericalangle BT_AA =\sphericalangle BAC $ Okay so now I will start my solution Solution- From the given condition we have that orthocenter $H$ of $\Delta ABC$ lies on $\odot(ADC)$, Now we will show that $X$ is the $B-HM point$ of $\Delta ABC$.

$X'\in \odot(BPD)$ and $X'\in BM$ so $X\equiv X'$ and we are done. Similarly we also have that $Y$ is the $D-HM point$ of $\Delta ADC$. Now we use property 3 to get that $XE$ and $YF$ are angle biectors of $\angle AXC$ and $\angle AYC$ Now we use a simple property VERY effectively , that the angle bisector passes through the midpoint of opposite arc of the circle. Now let $T_1, S_1$ be midpoints of arc $AC$ of $\odot (ABC)$ containing and not containing $B$ respectively. Analogously define $T_2, S_2$ for $D$. Now note that the above property gives 1. $PE$ passes through $S_1$ 2.$PF$ passes through $S_2$ 3. $QF$ passes through $T_1$ 4. $QE$ passes through $T_2$ Now if we fix $AC$ and the circles $\odot(ABC)$ and $\odot(ADC)$ then $T_1, S_1,T_2,S_2$ are fixed. And $P,Q$ are defined purely projectively in terms of $E,F$ , WLOG $E\neq M$ So now we fix $E$ and vary $F$ , we see that $P\rightarrow Q$ is a projectivity and as $E\rightarrow E$ it is a perspectivity. Now to show that the center is the point at infinity in direction perpendicular to $AC$ we just take $F$ as $M$ (getting $PQ$ as perpendicular bisctor of $AC$) and then take $F$ to be reflection of $E$ in $M$ ( here $PQ$ is line at infinity) Hence proved!!!

I guess this is what happens when you don't see the Humpty Points... Let $G=AB\cap CD,$ $H=AD\cap BC,$ and $L=D\infty_{AC}\cap \omega.$ Through an easy angle chase we find that $(GHBPD)=\omega.$ By the converse of Pascal's, $I=HE\cap GF$ is on $\omega$ since $GB\cap DH = A,$ $BP\cap HI=E,$ and $PD\cap IG=F$ are collinear. Applying the converse of Pascal's a second time, $J=DE\cap BF$ is also on $\omega$ since $BH\cap GD = C,$ $HI\cap DJ=E,$ and $IG\cap JB=F$ are collinear. Finally, using the converse of Pascal's a third time, $Q$ must also be on $\omega$ since $BX\cap YD=M,$ $XQ\cap DJ=E,$ and $YQ\cap BJ=F$ are collinear. Let $P'$ be the antipode of $P$ with respect to $\omega.$ Due to the facts that $-1=(A,C; M,\infty_{AC})\stackrel{D}{=}(H,G;Y,L)_{\omega}$ and $P$ is the arc midpoint of $\widehat{GDBH},$ we find that $YL,GG,HH,PP'$ must concur at some point $K.$ However, Brokard's on $BDGH$ implies that $AC$ must also pass through $K,$ and Pascal's on $PP'QYLD$ finishes the problem. $\blacksquare$

G6? Felt like a G2 or G3 ISL 2016 G6 wrote: Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$. Let $\overline{AD}\cap\overline{BC}=\{K\}$ and $\overline{AB}\cap\overline{CD}=\{T\}$. So, Obviously, $\angle PDK=\angle PBC\implies K\in\odot(BPD)$. Similarly $T\in\odot(BPD)$. Let $\overline{AC}\cap\odot(BPD)=\{U,V\}$. Then $(V,U;A,C)=-1\implies MU\cdot MV=MA^2=MX\cdot MB$. So, $\{X\}$ is the $A$-HM Point of $\Delta ABC$ and $Y$ is the $D$-HM Point of $\Delta ADC$. Let $\overline{XE}\cap\odot(BPD)=\{Z\}$.It's well known that $\odot(BXE)$ is the $B$- Appollonius Circle of $\Delta ABC$. Let $\odot(BXE)\cap\overline{AC}=X^*$. So, $\angle(\overline{PZ},\overline{AC})=\pi-\angle PZE-\angle ZEC=\pi-\angle XBE-\angle XEX^*=\pi-\frac{\pi}{2}=\frac{\pi}{2}$. Similarly if $\overline{YF}\cap\odot(BPD)=Z^*$. Then $\angle(\overline{PZ^*},\overline{AC})=\frac{\pi}{2}\implies Z\equiv Z^*\equiv Q$. So $\overline{PQ}\perp\overline{AC}$. $\blacksquare$

Let $H$ be the orthocenter of $ABC$. Note that $\angle AHC = 180^\circ - \angle ABC = 180^\circ - \angle ADC$, so $AHCD$ is cyclic. Redefine $X$ to be the $B$-HM point of $ABC$. We show that $X$ lies on $(APD)$, thereby showing that it is the same as the given $X$. We have, \begin{align*} \angle BPD &= 360^\circ - (\angle ABP + \angle ADP + \angle BAD) \\ &= (360^\circ - \angle ABC) - \angle BAD \\ &= (\angle BAD + \angle ADC + \angle DCB) - \angle BAD \\ &= \angle ABC + \angle DCB. \end{align*}Noting that $\overline{AC}$ is tangent to $(AXB)$ and that $X$ lies on $(BHC)$ (these are well-known properties of HM-points), we also have \begin{align*} \angle BXD &= \angle BXA + \angle AXD \\ &= (180^\circ - \angle BAC) + \angle ACD \\ &= (\angle ABC + \angle ACB) + \angle ACD \\ &= \angle ABC + \angle DCB. \end{align*}Hence, $BXPD$ is cyclic, meaning that the given $X$ is the $B$-HM point of $ABC$. Similarly, $Y$ is the $D$-HM point of $ADC$. Next, we claim that $\overline{XE}$ bisects $\angle AXC$. Indeed, consider an inversion about $(AC)$, which fixes $A$ and $C$ and swaps $X$ and $B$. This implies that $\triangle MAX \sim \triangle MBA$. In particular, $\frac{AX}{BA} = \frac{MX}{MA}$. Similarly, $\frac{CX}{CB} = \frac{MX}{MA}$, implying that $\frac{AX}{CX} = \frac{BA}{BC}$. As $\overline{AE}$ bisects $\angle ABC$, by angle bisector theorem $\frac{BA}{BC} = \frac{AE}{EC} = \frac{AX}{CX}$, implying that $\overline{XE}$ bisects $\angle AXC$ as claimed. Similarly, $\overline{YF}$ bisects $\angle AYC$. Next, we claim that $XQPY$ is cyclic. We have \begin{align*} \angle XPY &= \angle XPB + \angle BPD + \angle DPY \\ &= \angle XDB + \angle BPD + \angle DBY \\ &= (\angle XCA - \angle BDA) + (360^\circ - \angle ABC - \angle BAD) + (\angle ACY - \angle DBA) \\ &= \angle XBC + \angle CDY + \angle BAD - 180^\circ + (360^\circ - \angle ABC - \angle BAD) \\ &= \angle MBC + \angle CDM - \angle ABC + 180^\circ \\ &= (180^\circ - \angle BMC - \angle MCB) + (180^\circ - \angle DMC - \angle MCD) + (180^\circ - \angle ABC) \\ &= (\angle BMA - \angle MCB) + (\angle AMD - \angle MCD) + \angle AXC \\ &= \angle BMD - \angle BCD + \angle AXC. \end{align*}Additionally, \begin{align*} \angle XQY &= (360^\circ - \angle AXQ - \angle AYQ - \angle XAY) \\ &= (360^\circ - \angle AXC - \angle XAC - \angle YAC) \\ &= (360^\circ - \angle AXC - \angle ABX - \angle ADY) \\ &= (360^\circ - \angle ABM - \angle ADM) - \angle AXC \\ &= (\angle BAD + \angle BMD) - \angle AXC \\ &= \angle BMD + (360^\circ - 2\angle ABC - \angle BCD) - (180^\circ - \angle ABC) \\ &= \angle BMD + (180^\circ - \angle ABC) - \angle BCD \\ &= \angle BMD + \angle AXC - \angle BCD. \end{align*}Therefore, $XQPY$ is cyclic, so $AXQPYD$ is cyclic. To finish, \begin{align*} \angle(\overline{PQ}, \overline{BC}) &= 360^\circ - \angle AXQ - \angle XQP - \angle CAX \\ &= 360^\circ - \frac{\angle AXC}{2} - (180^\circ - \angle XDP) - \angle CAX \\ &= 180^\circ - \frac{180^\circ - \angle ABC}{2} + (\angle ADP - \angle ADX) - \angle CAX \\ &= 90^\circ + \frac{\angle ABC}{2} + \frac{\angle ABC}{2} - \angle ACX - \angle CAX \\ &= 90^\circ + (\angle ABC + \angle AXC) - 180^\circ \\ &= 90^\circ. \end{align*}We are done. $\Box$

Let $X_0$ be the intersection of $BM$ and $(ACD)$. Let $X'$ be the reflection of $X_0$ over $M$, which clearly lies on $(ABC)$ due to $(ABC)$ and $(ADC)$ being congruent and reflections over $M$. $~$ We have \begin{align*} \angle DX_0B &= \angle 180^\circ-\angle X'X_0D \\ &= 180^\circ - \angle X'X_0A +\angle XD_0A \\ &= 180^\circ - \angle CX'X_0 + \angle DCA \\ &= 360^\circ - \angle ADC - \angle DAC - \angle CAB \\ &= 360^\circ - \angle DAB - \angle PBA - \angle PDA \\ &= \angle DPB \end{align*}so $X_0=X$. Thus, $X$ lies on $(ACD)$ and similarly, $Y$ lies on $(ABC)$. $~$ Note that \[ \frac{CB}{CX} = \frac{\sin(\angle CXX')}{\sin(\angle CBM)} = \frac{\sin(\angle ACB)}{\sin(\angle CBM)} = \frac{MB}{CM} \]which is symmetric, and so we have $\tfrac{CB}{CX} = \tfrac{AB}{AX}$, which means that the feet of the angle bisectors of $\angle ABC$ and $\angle AXC$ concur on $AC$. That is, $XE$ bisects $\angle AXC$. Thus, letting $D_1$ and $D_2$ be the midpoints of arc $AC$ on $(ACD)$ with $D$, $D_1$ same side, and defining $B_1$ and $B_2$ similarly, $B_1B_2$ and $D_1D_2$ are the perpendicular bisector of $AC$. Also note that $D_1$ and $D_2$ lie on $DP$ and $XE$, respectively, due to angle bisector. \[\angle XQY = 180^\circ - \angle QD_1D_2 - \angle QB_1B_2 = 180^\circ - \angle XDP - \angle YBP = 180^\circ - \angle XDY\]so $Q$ is on $\omega$. Now, \[\angle QD_1B_1 = \angle XD_1B_1 = \angle X'B_1B_2 = \angle XBP = \angle D_1QP\]so $B_1D_1\parallel PQ$. Since $B_1D_1\perp AC$, we are done.

Let $K , L $ be the reflections of $B , D$ with respect to $M$, respectively. It's obvious that $K$ lies on $(ADC)$ and $L$ lies on $(ABC)$. Claim 1: $X , Y$ lies on $(ADC) , (ABC)$ respectively. Proof: Let $T$ be the intersection of lines $AL , DF$. $Y$ lies on $\omega$ is equivalent to $\angle BPT = \angle BYK$ which is equivalent to $\angle BAK = \angle BPT$ and since $\angle ABP = \angle ATP = \frac{\angle ABC}{2}$ , we have $\angle BAK = \angle BPT$. Thus $Y$ lies on $(ABC)$. Similarly $X$ lies on $(ADC)$. Claim 1 tells us $BY$ and $DX$ are the radical axes of $\omega , (ABC)$ ; $\omega , (ADC)$ respectively. Therefore $BY$ , $DX$ , $AC$ concurrent. Let $BY \cap DX = R$. Claim 2: $Q$ lies on $\omega$. Proof: Let $Q_1 = \omega \cap XE$. Apply Pascal's theorem on hexagon $XQ_1YBPD$ we have $E , R$ and $Q_1Y \cap DP$ are collinear. Therefore $Q_1Y \cap DP$ lies on $AC$ , in other words $Q_1Y \cap DP = F$ , so $Q_1 = Q$ as claimed. Since $MA \cdot MA = MA \cdot MC = LM \cdot MY = MY \cdot MD$ thus $\angle YAC = \angle ADY$, similarly $\angle YCA = \angle CDY$. By angle bisector theorem , we have $\frac{AY}{CY} = \frac{\sin(\angle ACY)}{\sin(\angle CAY)} = \frac{\sin(\angle CDY)}{\sin(\angle ADY)} = \frac{AD}{DC} = \frac{AF}{CF}$ so $YF$ is bisector of $\angle AYC$. Similarly $XE$ is bisector of $\angle AXC$. Let $M_1 = (ABC) \cap BE$ and $M_2 = (ADC) \cap DF$. $\angle QBD = \angle EXD = \angle EXA - \angle DXA = \frac{\angle AXC}{2} - \angle ACD = \angle PM_1M_2$ so $\angle QPM_1 = \angle QBD = \angle PM_1M_2$ so $PQ$ and $M_1M_2$ are parallel. Thus $PQ \perp AC$ as desired.

Let $K:=\overline{AD} \cap \overline{BC}$ and $L:=\overline{AB} \cap \overline{CD}$. Also define $W$ as the intersection of the tangents to $\omega$ at $K$ and $L$, and define $W'$ as the intersection of the tangents at $X$ and $Y$. Construct point $Z$ on $\omega$ such that $\overline{ZB} \parallel \overline{AC}$. In order to make the problem completely projective, we tweak the desired statement slightly by introducing the antipode $P'$ of $P$ on circle $\omega$, whence the problem asks to prove that $\overline{AC} \parallel \overline{QP'}$. Assume that $M$ is any point on $\overline{AC}$ until specified. After drawing a decent diagram, we see many undeniable observations, which we prove by ``Pascal spam." Claim: $W$ lies on $\overline{AC}$. Proof. Trivial by Pascal on $LLBKKD$. $\square$ Claim: $W'$ lies on $\overline{AC}$. Proof. Trivial by Brokard's theorem on $LKBD$ followed by $BDYX$. $\square$ Claim: [2016 G6'] $Q$ lies on $\omega$. By Pascal on $XKBYLD$, $\overline{XL} \cap \overline{YK} \in \overline{AC}$, and similarly Pascal on $LXDKYB$ implies that $\overline{XD} \cap \overline{YB} \in \overline{AC}$. Thus, the converse of Pascal theorem on $BPDXQY$ returns that $Q$ indeed lies on $\omega$. $\square$ Assume that $M$ is the midpoint of $AC$ henceforth. Then, $-1 = (AC;MP_{\infty})$, so projecting onto $\omega$, $(LK;XZ)$ is a harmonic bundle on $\omega$. By harmonic quadrilateral properties, this means that $W$, $X$, and $Z$ are collinear. Finally, doing Pascal on $ZBPP'QX$ finishes. $\blacksquare$

A proof featuring DIT for no reason, the inability to apply Pascal's, an inane homography, and a complex bash. Let $S = \overline{AB} \cap \overline{CD}$ and $T = \overline{AC} \cap {BD}$. Claim: $S$ and $T$ lie on $\omega$. Proof. We have that \[ \measuredangle DSB = \measuredangle SCB + \measuredangle CBS = \measuredangle DCB + \measuredangle CBP + \measuredangle PDC = \measuredangle DPB \]as desired. The proof for $T$ is symmetric. $\blacksquare$ Let $U = \overline{BF} \cap \omega$ and let $AC$ intersect the circle at $V, W$. Claim: $\overline{UD} \cap \overline{BP} = E$, so $F$ lies on the polar of $E$. Proof. By DIT on quadrilateral $STDB$ it follows that $(VW; AC) = -1$. As such, \[ (VW; AC) \overset{B}= (VW; ST) \overset{A}= (VW; BD) = -1 \]It thus follows that the polar of $DB$ lies on $AC$. Then, Brokard's on quadrilateral $PDUB$ gives that the polar of $\overline{DB} \cap \overline{UP}$ is $AC$, giving the result. $\blacksquare$ As such, by Converse Pascal on $XQYDUB$ (from which it follows that $E, F, M$ are collinear), it follows that $Q$ lies on $\omega$ as desired. Claim: $P$ lies on the perpendicular bisector of $ST$. Proof. Angle chase, then use $P$ lies on both internal bisectors. $\blacksquare$ Let $N$ be the other point on $\omega$ and the perpendicular bisector. We now rephrase the problem as follows. Projective Moment wrote: Let $TBDS$ be a cyclic quadrilateral in $\omega$ such that $A = \overline{TD} \cap \overline{BS}$ and $C = \overline{TB} \cap \overline{SD}$. Let $E$ and $M$ be a point on $AC$. Let the polar of $E$ intersect $AC$ at $F$. Define $P, X, Y, Q$ as originally. Let $\infty_M$ be the conjugate of $M$ wrt to $AC$. Let $N$ be the harmonic conjugate of $P$ wrt to $ST$. Let $O$ be the intersection of the polar of $\infty_M$ with $PN$, let $\infty_O$ be the harmonic conjugate of $PN$ with $O$. Let $H$ be the polar of $AC$. Show that $\overline{OH}, \overline{\infty_M\infty_O}$, and $\overline{PQ}$ concur. Map $TBDS$ to a rectangle. Then $H$ maps to the center of the polygon, $\overline{XQ} \parallel \overline{PB}$, and the reflection of $XB$ over $DS$ is perpendicular to $OH$. Then it remains to show that $OH$ bisects $\angle QAN$. Claim: The reflection of $Q$ over $OH$ is $N$. Proof. We now complex bash. Let $T, B, S, D$ be $-\overline{b}, b, \overline{b}, -b$ respectively. Let $X$ and $P$ be $x$ and $p$. Then $N$ is equal to $\overline{b^2p}$ and $T$ is equal to $pb\overline{x}$. Line $BX$ has slope equal to $b - x$, so $OH$ has slope equal to $i \cdot \overline{b - x}$. It remains to show that $- bx \cdot \overline{b-x}^2$ is a positive real number, which is well known. $\blacksquare$

No new points added??? Let $\angle ABC=\angle ADC=2x$. We begin with the following claim. Claim: $X$ is the $B$-humpty point of $\triangle ABC$. Similar relations hold for $Y$. Proof. Note that \[\angle XBA+\angle XDA=360 ^{\circ}- \angle BAD+\angle BQD= 360^{\circ} - \angle BAD-\angle BPD=\angle ABP+\angle ADP=2x\]Thus $\angle CDX=2x-\angle XDA=\angle XBA=\angle MBA$. Now, I claim that the only point on segment $BM$ that satisfies this is the humpty point. First, it works since $\angle CDX=\angle CAX=\angle MBA$ then. However, it is also the only point that works since $\angle CDX$ is fixed, and any point $K$ with $\angle CDK=\angle MBA$ is on one of two lines, the first of which contains $X$ and the humpty point and the second of which never intersects the interior of $ABCD$. This proves the claim since $X$ is on segment $BM$, which is in $ABCD$. $\blacksquare$ Now, we know that $\angle CXA=180^{\circ}-\angle CBA=180^{\circ}-\angle CDA$ so $ADCX$ and similarly $ABCY$ are cyclic. Thus, by radical axis on these two circles and $(BXPYD)$, we get that $BY, CX, AC$ concur, say at $L$. Claim: $Q$ is on $(PBD)$. Proof. Let $Q'=XE\cap (PBD)\neq X$. Then, pascal on $BPDXQ'Y$ gives $E,L, PD\cap Q'Y$ are collinear. This requires $PD\cap Q'Y=F$, giving $Q', F, Y$ collinear and $Q\equiv Q'$. $\blacksquare$ To finish, it's well known that $XE$ bisects $\angle AXC$, so $\angle AXQ=90-x$. Then, ignoring configuration issues, we have \begin{align*} 90 ^{\circ}&=\angle FDC+\angle AXQ=\angle FDC+\angle AXD+\angle QXD \\ &=(\angle FDC+\angle FCD)+(180^{\circ}-\angle QPF)=\angle QFE+(180^{\circ}-\angle QPF). \end{align*}This finishes as if $K=PQ\cap EF$, then $\angle FPK=180^{\circ}-\angle QPF$ so $\angle FPK+\angle PFK=90^{\circ}$ as desired.

?

Redefine $X$ and $Y$ as Humpty points in $\triangle ABC, \triangle ADC$ Since $MC^2 = MX \cdot MB = MY \cdot MD$ $\therefore BXYD$ is cyclic Claim 1:$P,Q \in (BXYD)$

Let $BB \cap AC = K$ $\measuredangle XQP + \measuredangle QEA = \measuredangle XBE + \measuredangle XEC= \frac{\measuredangle XKC + 180^\circ- \measuredangle XKC}{2}=90^\circ$.

Let $\omega_1$ be the circumcircle of $\triangle ABC$, and $\omega_2$ the circumcircle of $\triangle ADC$. Then $\omega_1$ is the reflection of $\omega_2$ over $AC$. Let $B'$ and $D'$ be the reflections of $B$ and $D$, respectively, over $AC$; they lie on $\omega_2$ and $\omega_1$ respectively. Now let segment $BM$ intersect $\omega_2$ at $X'$, and segment $DM$ intersect $\omega_1$ at $Y'$. Let $U$ be the midpoint of arc $AC$ on $\omega_1$ not containing $B$, and $V$ the midpoint of arc $AC$ on $\omega_2$ not containing $D$. Then $B$, $E$, $P$, and $U$ are collinear, $D$, $F$, $P$, and $V$ are collinear, and $V$, $M$, and $U$ are collinear. Extend $UV$ to intersect $\omega_1$ again at $W$ and $\omega_2$ again at $Z$. Let line $BM$ intersect $\omega_2$ at $S \neq X'$. Since \[ \angle BPD = \angle VPU = 180^\circ - \angle DVZ - \angle BUW = 180^\circ - \angle DVZ - \angle ZVS = 180^\circ - \angle DVS = 180^\circ - \angle DX'S = \angle BX'D, \]and similarly $\angle BPD = \angle BY'D$, $BX'PY'D$ is cyclic, and $(BX'PY'D) = \omega$. Therefore $X' = X$ and $Y' = Y$. Let $G$ be the radical center of $\omega$, $\omega_1$, and $\omega_2$, so $G$ lies on $AC$, $BY$, and $DX$. Now let $XE$ meet $\omega$ again at $Q'$. By Pascal's theorem on $XDPBYQ'$, $XD \cap BY = G$, $PB \cap XQ' = E$, and $DP \cap YQ'$ lie on one line, which must be $AC$. So $DP \cap YQ' \in AC$, so $DP \cap YQ' = DP \cap AC = F$, and $Y$, $F$, and $Q'$ are collinear. Since $Q'$ lies on both $XE$ and $YF$, $Q' = Q$. By reflecting $BU$ over $AC$, $B'$, $E$, and $V$ are collinear. Applying Pascal's theorem to $XZZVB'S$, we find that $XZ \cap VB'$, $ZZ \cap B'S$ (the point at infinity on line $AC$), and $XS \cap ZV = M$ lie on one line, which is $AC$. Since $XZ \cap VB' \in AC$, $XZ \cap VB' = AC \cap VB' = E$, and $E$ lies on $XZ$, giving $X$, $E$, $Q$, and $Z$ collinear. To finish, let $H = PQ \cap AC$. We have \[ \angle HPE = 180^\circ - \angle BPQ = 180^\circ - \angle BXQ = \angle SXZ = \angle SVZ, \]and \[ \angle HEP = \angle BEA = 180^\circ - \angle ABE - \angle EAB = 180^\circ - \frac12 \angle AB'C - \angle B'AC = 180^\circ - \angle VAC - \angle B'AC = 180^\circ - \angle VAB' = \angle VZS. \]Thus $\angle HPE + \angle HEP = \angle SVZ + \angle VZS = 90^\circ$, and $\angle EHP = 90^\circ$, giving $PQ \perp AC$.

$AD\cap BC=R, AB\cap DC=S$. Notice that $\angle DPB=360^{\circ}-\angle PBA-\angle BAD-\angle ADP=\widehat C+\frac{1}{2}\widehat D+\frac{1}{2}\widehat B=\widehat C+\widehat D=180^{\circ}-\angle DRB$ Hence $R,S\in \omega$. By Pascal on $XBBYDD$ and $RBBSDD$ we have $BY\cap DX\in AC$. By converse of Pascal on $XQYBPD$ we have $Q\in \omega$. $$\angle (QP,AC)=\angle (QP,PB)+\angle (PB,AC)=\angle (QX,XB)+\angle (BE,AC)=\angle (EX,XM)+\angle (BE,AC)$$Assume that $AC\cap \omega=\{T_1,T_2\}$. By harmonic bundles properties we have $MT_1.MT_2=MA.MC=MX.MB$, hence $X$ is fixed WRT $\triangle{ABC}$ when $D$ varies. Taking $D$ as the reflection of $B$ WRT $AC$ we see that $\angle EXM=\frac{1}{2}\widehat A-\frac{1}{2}\widehat C=90^{\circ}-\angle BEA$. $\blacksquare$

Let $\omega_b$ and $\omega_d$ denote the circumcircles of triangles $ABC$ and $ADC$, respectively. Since $\angle ABC=\angle ADC$, the two circles mentioned are reflections of each other across line $AC$. The main claim is the following : Claim : Points $X$ and $Y$ lie on $\omega_d$ and $\omega_b$, respectively. \newline Proof : Redefine $X$ and $Y$ to be the points where line segments $BM$ and $DM$ meet $\omega_d$ and $\omega_b$, so that $X$ is the $B$-humpty point in triangle $ABC$ and $Y$ is the $D$-humpty point in triangle $ADC$. We therefore have $\angle ADX=\angle ACX=\angle XBC$, where the last equality holds because the circumcircle of $(BXC)$ is tangent to line $(AC)$. Combining this with $\angle PBC=\angle PDA$ yields $\angle PBX=\angle PDX$, so $X$ indeed lies on $(BPD)$. By symmetry we also have that $Y$ lies on $(BPD)$, which establishes the claim. $\square$ We let $S_b$ and $N_b$ be the midpoints of the minor and the major arc $AC$ in $\omega_b$, respectively, and define $S_d$, $N_d$ similarly. Note that $S_b$ lies on line $\overline{B-P-E}$ and $S_d$ lies on line $\overline{D-F-P}$. Since $X$ is a humpty point, we have $\frac{XA}{XC}=\frac{BA}{BC}=\frac{EA}{EC}$. It follows by the angle bisector theorem that $(XE)$ must cut $\omega_d$ again at point $N_d$. Similarly, we get that points $Y$, $F$ and $N_b$ are collinear. Finally, note that $S_b$, $S_d$, as well as $N_b$, $N_d$ are reflection of each other across line $(AC)$. Point $M$ is thus the common midpoint of $S_bS_d$ and $N_bN_d$. Therefore, if $l_q$ is the line passing through $Q$ and perpendicular to $(AC)$, we have : \[-1=(N_b;N_d;M;\infty)\overset{Q}{=}(F;E;M;l_q\cap (AC))\]Defining $l_p$ similarly and projecting from $P$ points $S_d$, $S_b$, $M$ and $\infty$ also yields $-1=(F;E;M;l_p\cap (AC))$, which shows that $l_p$ and $l_q$ cut $(AC)$ at the same point and finishes the problem. $\square$

I omit all angle chasing because I solved this around $2-3$ months back and it's tiring to do all the work again. This is a kind of detailed outline from my notes.Some details might be extraneous because I found both the humpty points and some projective observations. Note that $(ABC)$ and $(ADC)$ are reflections of each other in $AC$. By some angle chasing, We show that $(BPD)$ passes through $AB \cap CD=G$ and $AD \cap BC=H$ . Now by radical axis theorem on $(BPD)$, $(ABC)$ and $(ADC)$:- We get that $BY$, $DX$ and $AC$ concur at a point which we call $T$. Now since $QX \cap PB=E$, $QY \cap DP=F$ and $BX \cap DY=T$ are all collinear, We get that by converse of pacal's on $XQYBPD$, all these points lie on a conic. Since $5$ points uniquely determine a conic and $5$ of these points already lie on a circle, we conclude that $Q$ also lies on this circle. Note that till here, we could've let $M$ be any point on $AC$ so that's $9$clubs for me on $OTIS$. Now some angle chasing shows that $X$ satisfies the same angle conditions which uniquely determine the $B$ humpty point of $\Delta ABC$ and similarly $Y$ is the $D$ Humpty Point of $\Delta ADC$. Now some more angle chasing, we conclude that $PQ \perp AC$.

I am doing the more general version, which says Quote: Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $\overline{AC}$ at $E$ and $F$ respectively, and intersect each other at $P$. Let $M$ be the any point on $\overline{AC}$ and let $\omega$ be the circumcircle of $\triangle BPD$, respectively. Segments $\overline{BM}$ and $\overline{DM}$ intersect $\omega$ again at $X$ and $Y$. Lines $\overline{XE}$ and $\overline{YF}$ meet at $Q$. Prove that $Q$ lies on $\omega$. We define only two new points. $\bullet$ $G=\overline{AB} \cap \overline{CD}$. $\bullet$ $H=\overline{AD} \cap \overline{BC}$. Claim: $G$, $H \in \omega$. Proof: Simple angle chase. Just see that \begin{align*} &\angle BPD=360^{\circ}-\angle BAD-\angle ABC\\ &\angle BGD=180^{\circ}-\angle GAD-\angle GDA=\angle ABC+\angle BAD-180^{\circ} \end{align*}And do the same for $\angle BHD$. $\blacksquare$ All of the synthetic work is done actually. We will now rephrase the problem in purely projective terms (we will even disregard one more Euclidean Condition now since it is not useful anymore). Quote: Let $HBDG$ be a quadrilateral inscribed in a circle $\omega$. Let $C=\overline{GH} \cap \overline{HB}$ and $A=\overline{HD} \cap \overline{GB}$. Let $M$ be a random point on $\overline{AC}$, and now introduce $X=\overline{MB} \cap \omega$, $Y=\overline{MD} \cap \omega$. Now let $P$ be a random point on $\omega$, and introduce $E=\overline{BP}\cap\overline{AC}$, $F=\overline{DP} \cap \overline{AC}$, $Q=\overline{YF} \cap \overline{XE}$. Prove that $Q$ lies on $\omega$. Now we take a homography sending $A$ to center of $\omega$ while preserving $\omega$, which means $HBDG$ is a rectangle, and $C$ becomes the point at infinity (the same direction as $\overline{HB}$). Now let $Q'=\overline{YF} \cap \omega$ and $E'=\overline{PB} \cap \overline{XQ'}$. Due to phantom point shenanigans, we just need to prove that $A$, $E'$, $C$ collinear. See that $\overline{XY} \parallel \overline{BD}$ and hence $\overline{XD}\cap \overline{YB}$ lies on $\overline{AC}$. Hence, just apply Pascal's on $YQ'XDPB$ to get that $F$, $E'$, $\overline{XD} \cap \overline{YB}$ are collinear, as desired.

Define $R,S=AD\cap BC,AB\cap CD$. Easy angle chase gives $R,S\in\omega$ and $PR=PS$. Define $P'$ the antipode of $P$ and we claim $Q$ lies on $\omega$ with $P'Q\parallel AC$. To show these, we solve the more general version: Quote: Let $RBDS$ be a quadrilateral inscribed in circle $\omega$ with $A=BS\cap DR,C=BR\cap DS$. Let $P,P'\in\omega$ with $(R,S;P,P')=-1$ and define $BP,DP\cap AC=E,F$. Let $M,N\in AC$ with $(A,C;M,N)=-1$ and define $BM,DM\cap\omega=X,Y$. Let $XE\cap YF=Q$. Show that $N,P',Q$ are collinear. Complex homography $RS\cap BD$ to the center of $\omega$. Now $RBSD$ is a rectangle, $P,P'$ are reflections over $RS$, and $X,Y$ is a diameter, and $Q$ satisfies $XQ\parallel BP,QY\parallel PD$ so $\angle XQY=\angle BPD=90^\circ$ so $Q\in\omega$. We want to show the bisectors of $\angle(P'Q,BX)$ are aligned to the sides of $RBSD$, which follows from an arc chase on $\omega$.

Proof that $Q$ lies on $\omega$ for any $M$

Rest of the proof

Cool problem but an exercise in homography. Denote by $S$ and $T$ the intersections of lines $AB$ and $CD$ and lines $AD$ and $BC$ respectively. We start off with the following observation. Claim : Points $B$ , $D$ , $S$ , $T$ and $P$ are concyclic (on say $\omega$). Proof : Note that, \[\measuredangle PDS = \measuredangle PBA\]which implies that quadrilateral $PDSB$ is cyclic. Similarly, quadrilateral $PDTB$ is also cyclic, which implies the claim. Let $Q=XE \cap \omega$. We wish to show that points $F$ , $Y$ and $Q$ are collinear. Now, take a homography which sends $BTDS$ to a rectangle while preserving its circumcircle (in particular we treat $P$ as a random point on $\omega$). [asy][asy] size(6 cm); import geometry; pair foot(pair P, pair A, pair B) { return foot(triangle(A,B,P).VC); } pair S = dir(150); pair B = dir(210); pair T = dir(330); pair D = dir(30); pair X = dir(100); pair Y = dir(280); pair P = dir(20); pair Q = dir(130); draw(S--B--T--D--cycle , blue+1.3); draw(Q--B,red); draw(P--X,red); draw(B--P,blue); draw(Q--X,blue); draw(Q--Y,blue); draw(D--P,blue); draw(Q--D,green+dashed); draw(P--Y,green+dashed); draw(circumcircle(D, B, S), blue); dot("$S$", S, dir(190)); dot("$B$", B, dir(B)); dot("$T$", T, dir(T)); dot("$D$", D, dir(20)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); [/asy][/asy] Note that since $\overline{AC}$ maps to the line at infinity, $BXDY$ is a rectangle. Further, $BQXP$ is an isosceles trapezoid so $QB=PX$. Thus, since we also have that $BD=ST$ and $\angle BQD = \angle XPY$ as $BD$ and $XY$ are both diameters of $\omega$, it follows that $\triangle BQD \cong \triangle XPY$. Thus, $QD=PY$ so $QDPY$ is an isoceles trapezoid. Thus, $DP \parallel QY$ and indeed points $F$ , $Y$ and $Q$ are collinear, which finishes the proof. Let $R$ denote the intersection of the perpendicular to $BP$ at $B$ and line $AC$. We can now note the following. Claim : Quadrilateral $BXRQ$ is cyclic. Proof : From the Right Angles/Bisectors picture we know that $(AC;ER)=-1$. Thus, \[ME\cdot MR = MC^2\]Further, it is well known that (and easy to check using linearity of power of point) \[MC^2 = \text{Pow}_{\omega}(M)=MB \cdot MX\]Thus, \[ME \cdot MR = MB \cdot MX\]and $BXRE$ must be cyclic. Denote by $Z$, the intersection of $PQ$ with line $AC$. Since $BXRE$ is cyclic we have, \[\measuredangle ZPB = \measuredangle QPB = \measuredangle QXB = \measuredangle EXB = \measuredangle ERB = \measuredangle ZRB\]Thus $ZPRB$ is cyclic. Finally we note, \[\measuredangle (PQ;AC) = \measuredangle PZE = \measuredangle PZR = \measuredangle PBR = \frac{\pi}{2}\]which finishes the problem.