Clearly, $B, P, Q, C$ lie on $\Gamma=(BC)$. Let $R=BP\cap CQ, S=BQ\cap CP$. Then $RSBC$ is an orthocentric system. The circle with diameter $RS$ passes through $P$ and $Q$ and is orthogonal to $\Gamma$. As there is only one circle through $P, Q$ orthogonal to $\Gamma$, it follows that $(RS)=\omega$. Let the tangent to $\omega$ at $S$ meet $MP$ at $X'$. Then $B, X', F$ are collinear along the polar of $C$ with respect to $\omega$, so $X=X'$. Similarly, $Y$ lies on the tangent to $\omega$ at $S$, so $XY$ is tangent to $\omega$ at $S$. [asy][asy] size(10cm); pair R=dir(90); pair S=dir(270); pair Q=dir(10); pair P=dir(220); pair B=extension(R, P, Q, S); pair C=extension(R, Q, P, S); pair M=midpoint(B--C); pair X = extension(P, P+dir(90)*P, S, S+dir(90)*S); pair Y = extension(Q, Q+dir(90)*Q, S, S+dir(90)*S); pair F= intersectionpoints(unitcircle, X--3*X-2*B, -1)[0]; pair E= intersectionpoints(unitcircle, Y--3*Y-2*C, -1)[0]; pair A = extension(B, E, C, F); draw(unitcircle); draw(circumcircle(B, P, C), dashed); draw(M--P); draw(M--Q); draw (B--R); draw(C--R); draw(B--Q); draw(C--P); draw(X--Y); draw(B--C); draw(A--B); draw(A--C); draw(B--F, dotted); draw(C--E, dotted); dot("$R$", R, dir(R)); dot("$S$", S, dir(S)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$M$", M, dir(270)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$F$", F, dir(F)); dot("$E$", E, dir(E)); dot("$A$", A, dir(90)); [/asy][/asy]

Dadgarnia wrote: In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$ be a circle inside of $ABC$ and is tangent to $AB,AC$ at $E,F$, respectively. The tangents from $M$ to $\omega$ meet $\omega$ at $P,Q$ such that $P$ and $B$ lie on the same side of $AM$. Let $X \equiv PM \cap BF $ and $Y \equiv QM \cap CE $. If $2PM=BC$ prove that $XY$ is tangent to $\omega$. Proposed by Iman Maghsoudi I will use the same diagram as a1267ab. All polars wrt $\omega$. Note that $P,Q$ lie on $\odot(BC)$. There exists a unique circle orthogonal to $\odot(BC)$ and passing through $P,Q$; precisely the circle $\odot(PSQR)$ where $S=\overline{BQ} \cap \overline{CP}$ and $R=\overline{BP} \cap \overline{CQ}$. Note that $S$ is the orthocenter of $\triangle RBC$. Thus, $\omega=\odot(PSQR)$. For complete quadrilateral $PSQRBC$ we note that $B$ lies on the polar of $C$. Consequently, $\overline{BF}$ is the polar of $C$. Thus, $X$ lies at the intersection of polars of $C$ and $P$; hence $X$ is the pole of $\overline{CP}$. Consequently, $\overline{XP}, \overline{XS}$ are tangents to $\omega$ since $S$ lies on both $\omega$ and $\overline{CP}$. Likewise, $\overline{YS}$ is tangent to $\omega$ at $S$; thus, $\overline{XY}$ is tangent to $\omega$ at $S$.

Honestly, the hardest part is to figure out how to draw an accurate diagram. After that, note that $\odot(BC)$ is orthogonal to $\omega$ so $B$ lies on polar of $C$ w.r.t. $\omega$. Now just pole-polar chase. The problem is equivalent to $BP\cap CQ\in\omega$ which is easy by angle chasing.

Once again, polars. Let $Z=\overline{BP}\cap\overline{CQ},\ S=\overline{PC}\cap\overline{BQ}$. The problem condition implies that $\angle BPC=\angle BQC=90^\circ$. Let $\omega'$ the circle with diameter $ZS$. Since $Z$ lies on the $S$-polar wrt. $(BPQC)$, $\omega'$ is orthogonal to $(BPQC)$. It is easy to prove that $S$ lies on $\omega$ (straightforward angle chasing) and $\omega$ is orthogonal to $(BPQC)$, too, hence $\omega=\omega'$ and $Z\in \omega$. Let $B',C'$ be the inverse points of $B,C$ wrt. $\omega$. We have $B',C'\in (BPQC)$. Since $\angle FC'C=\angle BC'C=90^\circ$, $BF$ must be the polar of $C$ wrt. $\omega$. Similarly, $CE$ is the $B$-polar. Observe that $P$ lies on the $C$-polar and $P$-polar, then $PC$ is the $X$-polar and analogously, $BQ$ is the $Y$-polar. Hence, $S$ is the pole of $XY$ wrt. $\omega$. Since $S\in \omega$, we must have $XY$ tangent to $\omega$.

The point $A$ is superfluous, so we don't include it in our diagram. Note that $MP=MQ=MB=MC$, so $BPQC$ is cyclic with diameter $BC$. Let $L=BP\cap CQ$. It is well known that $MP$ and $MQ$ are tangent to $(LPQ)$ given that $P$ and $Q$ are the feet of the altitudes from $B$ and $C$ in $\triangle LBC$. But the circle tangent to $MP$ and $MQ$ at $P$ and $Q$ respectively is unique, so we have $L\in\omega$. We are now essentially dealing with a standard triangle geometry problem with respect to $\triangle LBC$. Let $H$ be the orthocenter, let $D$ be the foot from $L$, and let $O$ be the midpoint of $LH$, i.e. the center of $\omega$. The big claim is that $C$, $B'$ and $E$ are collinear where $B'=OB\cap(BC)$, and similarly $B$, $C'$, and $F$ collinear. We see that $\omega$ and $(BC)$ are orthogonal, so $B'$ and $B$ are inverses in $\omega$. Furthermore, $\angle(OB,B'C)=\pi/2$, so the polar of $B$ with respect to $\omega$ is $CB'$, so $E$ is on $CB'$ since $E$ is the tangent from $B$ to $\omega$. The inverse of the tangent at $P$ to $\omega$ with respect to $\omega$ is $(OP)$, and the inverse of $BC'$ is $(OB'C)=(OC)$. Thus, the inverse of $X$, which is the intersection of these two objects must be the foot from $O$ to $CP$, or the midpoint of $PH$. Since $XP$ tangent to $\omega$, this clearly implies that $XH$ is also tangent to $\omega$. Similarly, $YH$ is tangent to $\omega$, so $XY$ tangent to $\omega$ at $H$. Remarks This problem is very similar in spirit to RMM 2013/3. In both, you add nice points to turn the problem into standard orthocenter geometry, and you prove a tangency by guessing the tangency point and then proving the tangency using standard methods.

[asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair M,B,C,P,Q,T,SS,EE,F,A,X,Y; path w; M=(0,0); B=dir(180); C=dir(0); P=dir(120); Q=dir(80); T=extension(B,Q,C,P); SS=extension(B,P,C,Q); w=circle( (T+SS)/2,length(T-SS)/2); EE=intersectionpoints(w,circle( (B+(T+SS)/2)/2,length(B-(T+SS)/2)/2))[0]; F=intersectionpoints(w,circle( (C+(T+SS)/2)/2,length(C-(T+SS)/2)/2))[0]; A=extension(B,EE,C,F); X=extension(M,P,B,F); Y=extension(M,Q,C,EE); draw(circle(M,1),gray); draw(Q--B--SS--C--P,gray); draw(B--F,gray+dashed); draw(C--EE,gray+dashed); draw(w); draw(X--Y); draw(P--M--Q); draw(A--B--C--A); dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,E); dot("$M$",M,S); dot("$P$",P,E); dot("$Q$",Q,W); dot("$T$",T,S); dot("$S$",SS,N); dot("$E$",EE,W); dot("$F$",F,NE); dot("$X$",X,W); dot("$Y$",Y,dir(30)); [/asy][/asy] First solution, by Brokard's theorem. Note that $B$, $P$, $Q$, $C$ all lie on the circle $\Gamma$ centered at $M$ with diameter $\overline{BC}$. Denote $S=\overline{BP}\cap\overline{CQ}$ and $T=\overline{BQ}\cap\overline{CP}$. Since $\angle BPC=\angle BQC$, we know $BCST$ is an orthocentric system, so by the Three Tangents lemma, $\overline{MP}$ and $\overline{MQ}$ are tangent to $(ST)$ at $P$ and $Q$ respectively. It follows that $\omega=(ST)$, so $\overline{ST}$ is a diameter of $\omega$. By Brokard's theorem, $B$ is the pole of $\overline{CE}$ and $C$ is the pole of $\overline{BF}$ with respect to $\omega$. It follows that $X$ is the pole of $\overline{CP}$, so $\overline{XT}$ is tangent to $\omega$. Similarly $\overline{YT}$ is tangent to $\omega$, so $\overline{XY}$ is tangent to $\omega$ at $T$, as desired. Second solution, by self-polar orthogonality. Note that $B$, $P$, $Q$, $C$ all lie on the circle $\Gamma$ centered at $M$ with diameter $\overline{BC}$. Thus $\omega$ and $\Gamma$ are orthogonal, so by self-polar orthogonality, $B$ and $C$ lie on the polars of each other with respect to $\omega$. This implies $X$ is the pole of $\overline{CP}$ and $Y$ is the pole of $\overline{BQ}$. It thus suffices to show that $T=\overline{BQ}\cap\overline{CP}$ lies on $\omega$, but this is just angle-chasing: let $N$ be the center of $\omega$; then, \[\angle PTQ=\frac{\widehat{BC}+\widehat{PQ}}2=90^\circ+\frac12\angle PMQ=180^\circ-\frac12\angle PNQ,\]and the conclusion follows from inscribed angle theorem.

Iran TST 2018 Day 1 P3 wrote: In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$ be a circle inside of $ABC$ and is tangent to $AB,AC$ at $E,F$, respectively. The tangents from $M$ to $\omega$ meet $\omega$ at $P,Q$ such that $P$ and $B$ lie on the same side of $AM$. Let $X \equiv PM \cap BF $ and $Y \equiv QM \cap CE $. If $2PM=BC$ prove that $XY$ is tangent to $\omega$. Proposed by Iman Maghsoudi From the condition $2PM=BC$ it is clear that $P,Q$ lies on the circle centered at $M$ with diameter $MB$. Let this circle be $\gamma$. So, $\omega\perp\gamma$. We restate the problems in terms of $\omega$ and $\gamma$. Rephrased Problem wrote: Let $\gamma$ be a circle with center $M$ and a diameter $BC$, and let $\omega$ be a circle orthogonal to $\gamma$. Let the tangents from $B,C$ be tangent to $\omega$ at $E,F$ respectively meet at a point $A$. Let $\{\omega\cap\gamma\}=\{P,Q\}$. If $\{BF\cap MP,CE\cap QM\}=\{X,Y\}$ respectively. Then $XY$ is tangent to $\omega$ Note that as $\omega\perp\gamma$ and $CB$ is the diameter of $\gamma$. By Self Orthogonality Lemma we get that $B\in\text{ Polar of } C$ WRT $\omega$ and vice versa (By La Hire's). So, if $\{CE,BF\}\cap\{\omega\}=\{K,T\}$ respectively then $BK,CT$ are tangent to $\omega$. Now as $X\in\text{ Polar of } C$, so by La Hire's Theorem we get that $C\in\text{ Polar of } X$. So, if $CP\cap\omega=R$, then $XR$ is tangent to $\omega$. But $\overline{Q-R-B}$ (as it is well known that the circle with diameter with endpoints $PB\cap QC$ and $QB\cap PC$ forms an orthocentric system with $\gamma$). So, again by La Hire we get that $B$ lies on Polar of $Y\implies QB$ is the Polar of $R$ WRT $\omega$. So, $YR$ is also tangent to $\omega$. So, $XY$ is tangent to $\omega.$ $\blacksquare$

Note that $(BPQC)$ is cyclic with center $M.$ Moreover, the given conditions imply that $\omega$ is orthogonal to $(BC)$, so by Self Polar Orthagonality $B$ and $C$ are polar with respect to each other in $\omega.$ Now, it follows that $BP \cap CQ = R$ lies on $\omega$, as does $BQ \cap CP = S.$ Claim: $X, S$, and $Y$ are collinear Proof: Suppose that $T$ is the intersection of $CE$ and $BF.$ It follows that $T$ is polar to both $B$ and $C$, since $CE$ is the polar of $B$ and $BF$ is the polar of $C.$ Therefore, $T$ is also the intersection of $PQ$ and $RS$ by Brokard, so applying Pappus on $\overline{PTQ}$ and $\overline{BMC}$ yields the desired result. To finish, we will show that $SY$ is tangent to $\omega$, which finishes the problem since analogous reasoning will yield $SX$ tangent to $\omega$. Let the tangent $SS$ intersect $QM$ at $Y'$. Then, apply Pascal to $SSPQQR$, meaning $SS \cap QQ = Y'$, $SP \cap QR = C$, and $PQ \cap RS = T$ are collinear, implying $Y = Y'$ as desired.

motivated by the well-known lemma in $\triangle{ABC}$ $ME$ is tangent to $AEF$ where $M$ is the midpoint of $BC$ and $E,F$ are the foot of altitudes from $B,C$ define $H=CP \cap BQ$ and $D=BP\cap CQ$ we have $MP=MQ=MB=MC$ so $\angle{BQC}=\angle{BPC}=90$ so $H $ is the orthocenter of $\triangle{DBC}$ so from the lemma above $D,H \in \omega$ from brokard on $DPHQ$ we have $C \in Polar(B)$ wrt $\omega$ so $Y \in Polar(B)$ and clearly $Y \in Polar(Q)$ so $Y \in Polar(H)$ so $YH$ is tangent to $\omega $ similary $XH$ is tanget to $\omega$ and so $XY$ is tangent and we win

Since $2PM=BC$ we have $B,P,Q,C$ all lies on a circle with centre $M$. Denote this circle by $\gamma$. Now $\gamma$ and $\omega$ are orthogonal. This eliminates $A$. CLAIM 1. $CE$ is the polar of $B$ w.r.t. $\omega$. Proof. Obviously $E$ lies on the polar of $B$. Let $BI$ meet $\gamma$ again at $K$. Then since $IK\times IB=IP^2=IE^2$, $K$ lies on the polar of $B$. Since $\angle EKB=\angle IEB=\angle BKC=90^{\circ}$, $E,K,C$ are collinear, we establish the claim. Now by symmetry $BF$ is the polar of $C$ w.r.t. $\omega$. CLAIM 2. Suppose $BQ$ and $CP$ meet at $T$, $BP$ and $CQ$ meet at $U$. Then $U$ and $T$ are antipode w.r.t. $\omega$ Proof. Since $\angle UPC=\angle BPC=\angle BQC=\angle UQB=90^{\circ}$, $U,P,T,Q$ are concyclic. Moreover, let $PQ$ and $BC$ meet at $V$. By Brokard's theorem $UT$ is the polar of $V$ w.r.t. $\gamma$. By La Hire's theorem $I$ also lies on the polar of $V$ w.r.t. $\gamma$. This shows that $U,I,T$ are collinear. This establishes CLAIM 2. Now by claim 1, $X$ lies on the polar of $C$ w.r.t. omega, hence $C$ lies on the polar of $X$. Hence $XT$ is tangent to $\omega$. Similarly $YT$ is tangent to omega. This shows that $XY$ is tangent to $\omega$.

Let $H= CP \cap BQ, Z= BP \cap CQ$ $\implies$ since $MB=MP=MQ=MC \implies M$ is the center of $(BPQC)$ $\implies$ $BC$ is diameter of $(BPQC) \implies$ $PZHQ$ is cyclic with diameter $ZH$ and $H$ is the ortocenter of $\Delta ZBC \implies$ it's well known that $MP,MQ$ touch $(PZHQ)$, and since $MP,MQ$ also touch $\omega$ at $P,Q$, we have that $(PZHQ)= \omega$. Now, suppose that the tangent to $\omega$ through $H$ intersects $MP,MQ$ at $X',Y'$, respectively, and $E'=CY' \cap \omega, F'=BX' \cap \omega$, with $E',F' \neq X',Y'$ and $E',F'$ outside of $\Delta ZBC$. By Brokard's Theorem, since $B=HQ \cap ZP \implies B \in \Pi_{\omega}(C)$. $(*)$ Also, since $C$ lies on $PH= \Pi_{\omega}(X')$, by La Hire's Theorem, $X' \in \Pi_{\omega}(C) \implies$ from $(*)$, $BX'= \Pi_{\omega}(C)$. Thus, since $F'=BX' \cap \omega \implies F' \in \Pi_{\omega}(C) \implies CF'$ touches $\omega$ $\implies F'=F$. Similarly, $E'=E \implies X'=X,Y'=Y \implies XY$ is tangent to $\omega$ at $H$ $\implies XY$ is tangent to $\omega$, as desired. $\blacksquare$

Since $2MP = BC$ $\Rightarrow MP = MQ = MB = MC$ $\Rightarrow P, Q$ lies on circle with diameter $BC$. We notate $(XY)$ which means circle with diameter $XY$. Let $CP \cap BQ = S$, $BP \cap CQ = K$ and $I$ be the midpoint of $KS$. Clearly $BS \perp CQ$ and $CS \perp CP$, so $S$ is the orthocenter of $\triangle KBC$, hence $(KS)$ is orthogonal to $(BC)$. On the other hand, $\omega$ is also orthogonal to $(BC)$ (since $MP, MQ$ are tangents to $\omega$ at $P, Q$ and $P, Q \in (BC)$). That gives us $(KS) \equiv \omega$. So $B, C$ are conjugate wrt $\omega$, also $B, E$ are conjugate wrt $\omega$ ($BP$ is tangent to $\omega$ at $E$) $\Rightarrow$ $EC$ is the polar of $B$ wrt $\omega$. Then with $Y \in EC$, the polar of $Y$ wrt $\omega$ passes through $B$ (La Hire theorem), notice $YQ$ is tangent to $\omega$ at $Q$ $\Rightarrow BQ$ is polar of $Y$ wrt $\omega$. So with $BQ \cap \omega = S$ then $YS$ is tangent to $\omega$ at $S$. From here we have angle transformation: $\angle YSQ = \angle QKS = \angle QBC$ Hence $YS \parallel BC$. With similar argument, we also get $XS \parallel BC$. That gives us $\overline{X, Y, S}$ and $XY$ is tangent to $\omega$ at $S$. $\blacksquare$

$Lemma1)$.Let $(ABC)$ be a triangle $(ABC)$ and $D$,$E$,$F$ its foot of perpendiculars from $A$,$B$,$C$ respectively.Let $AD$ intersect $EF$ at $Q$.Now let $BQ$ and $CQ$ intersect the circle of $(AEF)$ at $X$,$Y$ respectively.Prove that $CX$ and $BY$ are tangent to $(AEF)$. We prove this using pole-polars with respect to $(AEF)$.From Brocard theorem on $(AFHE)$ we have that $C$ is the pole of $BQ$, similarly $B$ is the polar of $CQ$. $T$ is on $BQ$ ,so the polar of $X$ will pass through the pole of $BQ$ which is $C$, so we get that $CX$ isw tangent to $(AFE)$, similarly we get that $BY$ is tangent to $(AEF)$. Now we get to our problem, Let tangents to $H$ and $F$ and to $H$ and $E$ intersect at $P$ and $R$.We are going to porve that $B-P-Q$ and $C-R-Q$ collinear. So the polar of $P$is $FH$, since $C$ lies on $EF$ we get that the polar of $C$ will pass through the pole of $HF$, so this means that $BQ$ will pass through $P$.Which means that $B-P-Q$ collinear and we get that $B-P-Q-X$ collinear Similarly we get that $C-R-Q-Y$-collinear. $\blacksquare$

dame dame

Well it's almost the same essence compared to the above post. Thanks to Aayam for his hint! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.171371144929623cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.036719358139617, xmax = 8.134651786790007, ymin = -4.178407314126202, ymax = 4.820610351447039; /* image dimensions */ pen dqfqcq = rgb(0.8156862745098039,0.9411764705882353,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-4.89,4.12)--(-6.59,-3.4)--(2.79,-3.28)--cycle, linewidth(0.4) + dqfqcq); /* draw figures */ draw((-4.89,4.12)--(-6.59,-3.4), linewidth(0.4) + dqfqcq); draw((-6.59,-3.4)--(2.79,-3.28), linewidth(0.4) + dqfqcq); draw((2.79,-3.28)--(-4.89,4.12), linewidth(0.4) + dqfqcq); draw(circle((-3.9968484812260634,1.135908479816047), 1.5291597834262822), linewidth(0.4) + wrwrwr); draw(circle((-1.9,-3.34), 4.690383779606952), linewidth(0.4) + wrwrwr); draw((-5.106197920323259,0.08344780823584408)--(2.79,-3.28), linewidth(0.4) + wrwrwr); draw((-6.59,-3.4)--(-2.4781667183844123,1.3146131145082935), linewidth(0.4) + wrwrwr); draw((-5.106197920323259,0.08344780823584408)--(-2.4781667183844123,1.3146131145082935), linewidth(0.4) + wrwrwr); draw((-6.59,-3.4)--(-4.043161310149092,2.668373296746161), linewidth(0.4) + wrwrwr); draw((-4.043161310149092,2.668373296746161)--(2.79,-3.28), linewidth(0.4) + wrwrwr); draw((-3.950535652303035,-0.3965563371140669)--(-4.043161310149092,2.668373296746161), linewidth(0.4) + wrwrwr); draw((-5.106197920323259,0.08344780823584408)--(-1.9,-3.34), linewidth(0.4) + wrwrwr); draw((-1.9,-3.34)--(-2.4781667183844123,1.3146131145082935), linewidth(0.4) + wrwrwr); draw((-4.648294463719957,-0.4054828890724923)--(-2.2682892662635625,-0.375035061386483), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-4.89,4.12),dotstyle); label("$A$", (-4.836644944646243,4.257268956651925), NE * labelscalefactor); dot((-6.59,-3.4),dotstyle); label("$B$", (-6.526669129031589,-3.2539496406162707), NE * labelscalefactor); dot((2.79,-3.28),dotstyle); label("$C$", (2.847909466405077,-3.138392431427529), NE * labelscalefactor); dot((-1.9,-3.34),linewidth(4.pt) + dotstyle); label("$M$", (-1.8466021568875526,-3.2250603383190852), NE * labelscalefactor); dot((-5.488371109628718,1.473087797407073),linewidth(4.pt) + dotstyle); label("$E$", (-5.428875641738544,1.5850084941622782), NE * labelscalefactor); dot((-2.9358289132634257,2.2370747341340276),linewidth(4.pt) + dotstyle); label("$F$", (-2.8721723884376344,2.3505750050376903), NE * labelscalefactor); dot((-5.106197920323259,0.08344780823584408),linewidth(4.pt) + dotstyle); label("$P$", (-5.053314711875133,0.19832198389738068), NE * labelscalefactor); dot((-2.4781667183844123,1.3146131145082935),linewidth(4.pt) + dotstyle); label("$Q$", (-2.4243882028312607,1.4261173315277587), NE * labelscalefactor); dot((-3.950535652303035,-0.3965563371140669),dotstyle); label("$J$", (-3.8977426199877168,-0.2494622017089925), NE * labelscalefactor); dot((-4.043161310149092,2.668373296746161),dotstyle); label("$J'$", (-3.984410526879273,2.812803841792656), NE * labelscalefactor); dot((-3.9890929905502968,0.6122662132644551),linewidth(4.pt) + dotstyle); label("$K$", (-3.926631922284902,0.7327740763953099), NE * labelscalefactor); dot((-4.648294463719957,-0.4054828890724923),linewidth(4.pt) + dotstyle); label("$X$", (-4.591085875120166,-0.29279615515477053), NE * labelscalefactor); dot((-2.2682892662635625,-0.375035061386483),linewidth(4.pt) + dotstyle); label("$Y$", (-2.20771843560237,-0.2639068528575852), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Note that $BPQC$ is cyclic. Now define $$ J=PC\cap BQ,~~J'=PB\cap CQ.$$ Claim: $PJ'QJ$ is $\omega.$ To show this it's enough to show that, $PM,~~PQ$ tangent to it at points $P,Q.$ Note that $$\angle JPM=\angle PCM=\angle PQB\implies PM\text{~~is tangent to~~} J'PJQ\text{~~at~~} P. $$Similarly we get $QM$ is tangent to $J'PJQ$ at $Q.$ Hence $PJ'QJ$ is $\omega.$ Now, consider the parallel line through $J$ to $BC.$ Let it intersect $MP,~~MQ$ at $X',Y'$ respectively. Note that $$ \angle PJ'X'=\angle PCM=\angle MPC=\angle X'PJ\implies X'PJ\text{~~is isosceles~~}$$$$\implies XJ'\text{~~is tangent to}\omega.$$Similarly, we get $Y'JQ$ isosceles $\implies Y'J $ is tangent to $\omega.$ Angle chase gives us $X'-J-Y'.$ Hence if we show $B-X'-F$ and $C-Y'-E,$ we will be done. Claim: $B-X'-F$ are collinear Note that $B-K$ is the polar of $C$ by brokards. ( where $K=PQ\cap JJ'$) Also, note that $C\in $ polar of $X'$ wrt $\omega.$ So by la-hire, we get $X'$ lying in the polar of $C.$ But we also know that $F$ is the tangency point of $C$ to $\omega.$ Hence $F$ lies in the polar of $C.$ Hence $B-X'-K-F.$ Similarly, we get $C-Y'-K-E.$ Hence $X'=X,~~Y'=Y.$ And we are done!

Kudos to those who didn't use geogebra! Dadgarnia wrote: In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$ be a circle inside of $ABC$ and is tangent to $AB,AC$ at $E,F$, respectively. The tangents from $M$ to $\omega$ meet $\omega$ at $P,Q$ such that $P$ and $B$ lie on the same side of $AM$. Let $X \equiv PM \cap BF $ and $Y \equiv QM \cap CE $. If $2PM=BC$ prove that $XY$ is tangent to $\omega$. Proposed by Iman Maghsoudi Construct a circle with diameter $BC$,note that $\odot(BC)$ and $\omega$ are orthogonal. Say $BP\cap CQ=H$ and $BQ \cap CP=I$ Clearly $PHQI$ is cyclic ! Claim : $\odot(PHQI) \equiv \omega$ Proof : By angle Chasing, $$\angle MPI=\angle MPC=\angle MCP=\angle BCP=\angle BQP \implies \text{MP is tangent to (PHQI) !}$$Similarly $MQ$ is also tangent to $(PHQI)$ Thus our claim is proved $\blacksquare$ Now by self polar orthogonality $BF$ is the polar of $C$ and combining this with La-Hire's theorem we get that $C$ lies on the polar of $X$. Again as $X \in MP\implies XP$ is tangent to $\omega$,this implies $PC$ is the polar of $X$.Similarly $BQ$ is the polar of $Y$.Finally we again apply La-Hire's theorem to get that $BQ \cap CP=I$ is the pole of $XY$.But by our previous claim we know $I\in \omega$,thus $XY$ is tangent to $\omega \text{ } \blacksquare$

All poles and polars in this proof will be with respect to $\omega$. Now we define some new points. $\bullet$ $T=\overline{PC} \cap \overline{QB}$. $\bullet$ $R=\overline{EC} \cap \overline{FB}$. The condition in the problem statement is equivalent in saying that the circle with diameter $\overline{BC}$ is orthogonal to $\omega$ and hence $B$ and $C$ lie on each others polars; and hence, $C$ has polar $\overline{BF}$ and $B$ has polar $\overline{CE}$. Claim: $T$ is the pole of $\overline{XY}$. Proof: See that $X \in$ polar of $C$ and $Y \in$ polar of $B$, which means polar of $X$ is $\overline{PC}$ and polar of $Y$ is $\overline{QB}$ and hence $T$ has polar $\overline{XY}$ (see that we used La Hire a bunch of times here). $\blacksquare$ Claim: $P$, $Q$, $R$ is collinear. Proof: See that polar of $\overline{PQ}$, $\overline{CE}$, $\overline{BF}$ are concurrent and now just take the dual statement. $\blacksquare$ Claim: $T$, $X$, $Y$ are collinear. Proof: Apply Pappus' Theorem on $PRQ$ and $BMC$. $\blacksquare$ Obviously we are done now.

Great problem Let $I$ be the center of $w$ and let $T=BQ\cap CP$, which we claim is the desired intersection point. Define the linear function $f(X)=\mathbb{P}(X,M)-\mathbb{P}(X,w)$. $$f(B)+f(C)=2f(M)\Rightarrow -BM^2+BE^2-CM^2+CF^2=2MP^2\Rightarrow$$$$ BE^2+CF^2=BC^2\Rightarrow BE^2+CI^2=BC^2+EI^2$$Thus by the perpendicular lines lemma $BI\perp CE$. As $E$ lies on the polar of $B$ we must then have that $C$ also lies on the polar of $B$ and vice versa. Then $$\angle PTQ=\angle TCQ+\angle TQC=\frac{1}{2}\angle PMQ+90^{\circ}=180^{\circ}-\frac{1}{2}\angle PIQ$$so we must have $T\in w$. As the polar of $BF$ lies on $PT$ we must have that the polar of $PT$ lies on $BF$ or that $TX$ is tangent to $w$. As $TY$ is also tangent to $w$, we are done.

Literally no angle chasing required??????????????? I will use @a1267ab's diagram. Evidently $BPQC$ is cyclic, so $S=PC\cap BQ$ lies on $\omega$. It suffices to show that the polar of $S$ (all pole/polars wrt $\omega$) is $XY$. By La Hire's, we just need to show that the polar of $X$ (and similarly $Y$) goes through $S$. Now, as $X=PM\cap BF$, we just need to show that the poles of lines $PM$ and $BF$ are collinear with $S$, by La Hire's again. The pole of line $PM$ is obviously $P$. The pole of $BF$ is $C$ beacuse $CF$ is tangent to $\omega$ and $C$ lies on the polar of $B$ by Brokard's theorem. Thus, we are done.

Solved with Narwhal234. I really like how a random lemma I learnt nearly two years ago suddenly came to my rescue. It is clear that the condition $MP = \frac{1}{2}BC$ translates to the fact that $P$ and $Q$ lie on the circle with diameter $BC$, we denote this circle by $\gamma$. We first make a preliminary observation about these two circles. Claim : Circles $\gamma$ and $\omega$ are orthogonal. Proof : Let $O$ denote the center of $\omega$. In triangles $\triangle OPM$ and $\triangle OQM$, we have $MP=MQ$ and $OP=OQ$. Further $OM$ is a common side implying that $\triangle OMP \cong \triangle OMQ$. Thus, \[\measuredangle OPM = \measuredangle MQO = \frac{\pi}{2}\]which implies the claim. Now, we approach the problem as follows. Let $R = \overline{BQ} \cap \overline{CP}$. Now note that, \[\measuredangle PRQ = \frac{\pi}{2} + \measuredangle PBQ = \frac{\pi}{2}+ \measuredangle PMO = \measuredangle MOP \]which implies that $R$ lies on $\omega$ as well. Now, we attack the problem. Claim : Points $X$ and $Y$ lie on the polar of $R$ with respect to $\omega$. Proof : Note that since $\omega$ and $\gamma$ are orthogonal and $BC$ is a diameter of $\gamma$, by the Self Polar Orthogonality Lemma, it follows that $C$ lies on the polar of $B$ with respect to $\omega$. Next, since clearly $BE$ is a tangent to $\omega$, this implies that $\overline{CE}$ is the polar of $B$ with respect to $\omega$. Thus, $Y$ lies on the polar of $B$ with respect to $\omega$ as well. By La Hire's this implies that $B$ lies on the polar of $Y$ with respect to $\omega$. Also clearly $YQ$ is a tangent to $\omega$ so $Q$ must lie on the polar of $Y$ with respect to $\omega$. This implies that $\overline{BQ}$ is the polar of $Y$ with respect to $\omega$. Thus, $R$ lies on the polar of $Y$ with respect to $\omega$. By La Hire's again this implies that $Y$ lies on the polar of $R$ with respect to $\omega$, as claimed. A similar argument shows that $X$ also lies on the polar of $R$ with respect to $\omega$. Now since $R$ is a point on $\omega$, its polar with respect to $\omega$ is simply the tangent to $\omega$ at $R$. Thus, $\overline{XY}$ is the tangent to $\omega$ at $R$ which solves the problem.

cool problem! had to use ggb but it was mostly just La-Hire's