Hint: consider the circle passing through $M$ and $P$ which is tangent to $AB$ and $AC$.

Let $\ell: A\in\ell$ is the line, parallel $BC$. Consider the projective transformation, which maps $\ell$ in the infinite line and preserve insircle of $ABC$. Then ratios of distances on the any line, parallel $\ell$, are preserved. Then the problem is trivial: the points $P'$ and $Q'$ in which maps the points $P$ and $Q$ are symmetric of the center of incircle.

PP.Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. Let $\gamma$ be the incircle of triangle $ABC$. The median $AM$ of triangle $ABC$ intersects the incircle $\gamma$ at two points $K$ and $L$. Let the lines passing through $K$ and $L$, parallel to $BC$, intersect the incircle $\gamma$ again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $P$ and $Q$. Prove that $BP = CQ$. Lemma. Let $ABC$ be a triangle for which $b\ne c$. Denote: the middlepoint $M$ of the side $[BC]$; the points $D,E,F$ where the incircle $C(I,r)$ touches the side-lines $BC,CA,AB$ respectively; the intersections $K,L$ between the incircle and the median $AM$; the intersection $S\in EF\cap AM$. Then $I\in SD$ and the division $(A,K,S,L)$ is harmonically. Proof. $\frac{FB}{FA}\cdot MC+\frac{EC}{EA}\cdot MB=\frac{SM}{SA}\cdot BC\Longrightarrow\frac{SM}{SA}=\frac{a}{b+c-a}\ \ (1)$. Therefore, $T\in BC,\ AT\perp BC\Longrightarrow $ $MD=\frac 12\left| b-c\right|\ ,$ $DT=$ $\frac{(p-a)|b-c|}{a},\ \frac{DM}{DT}=\frac{a}{b+c-a}\ \ (2)$. From the relations $(1)$ and $(2)$ results that $\frac{SM}{SA}=\frac{DM}{DT}$, i.e. $SD\parallel AT$ what means $I\in SD$. Immediately we have and the second part of the conclusion (the line $EF$ is the polar of the point $A$ w.r.t. the incircle). Proof of the proposed problem. Define the point $S\in EF\cap AM$ . Observe that $KX\parallel LY\iff$ $KXLY$ is an isosceles trapezoid $\iff$ $S\in XY\cap KL\implies$ $\frac {AK}{AL}=\frac {SK}{SL}=\frac {KX}{YL}\implies$ $\boxed{\frac {AK}{AL}=\frac {KX}{YL}}\ (1)$ .Therefore, $\left\{\begin{array}{cccc} \frac{AM}{AK} & = & \frac{MP}{KX} & (2)\\\\ \frac{AL}{AM} & = & \frac{LY}{MQ} & (3)\end{array}\right\|$ . From the product of the relations $(1)$ , $(2)$ and $(3)$ we obtain $MP=MQ$, i.e. $BP=CQ\ .$

Very nice!

Thanks, Arne ! I and you must at first prove that the points $A,K,Z,L$ form a harmonical division. This property is one among the first problems in which the my mathe-teacher applied the harmonical division (approx. in 1958 !)

What does harmonic division mean ? Virigil please can you send a link or something like that ... Davron

Davron and Arne, see that (in the previous message) I proved the mentioned classical harmonical division in a lemma.

A less sophisticated solution, using Arne's hint:

My solution is different:

Generalization of this problem : Theorem Let $ABC$ be a triangle, and $P, Q$ is the point on $BC$ such that $BP=CQ$. Let $w$ be the incircle of triangle ABC. The line $AP$ intersects the incircle $w$ at $R$ (closer to A) and line $AQ$ intersects the incircle $w$ at $S$ (farrer to A). Let the lines passing through $R$ and $S$, parallel to BC, intersect the incircle w again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $Z$ and $W$. than $BZ=CW$ it is modification of India 1998 and easy to prove. we can use trigonometry or power of point. Sorry to my bad english.

Virgil Nicula wrote: $\frac{FB}{FA}\cdot MC+\frac{EC}{EA}\cdot DB=\frac{SM}{SA}\cdot BC$. Well, why is this?

Hmm... I still don't understand. What theorem did you use here?

Quote: Lemma. Let be given a triangle $ABC$ and a point $M$ lies on segment $BC.$ A line cuts $AB,\,AC,\,AM$ at $B',\,C',\,M'$ respectively. We have: \[BC\cdot\frac{AM}{AM'}=BM\cdot\frac{AC}{AC'}+CM\cdot\frac{AB}{AB'}\] Proof. We have: \begin{eqnarray*}\frac{AB'\cdot AC'}{AB\cdot AC}&=&\frac{S(AB'C')}{S(ABC)}\\&=&\frac{S(AB'M')+S(AC'M')}{S(ABC)}\\&=&\frac{S(AB'M')}{\frac{BC}{BM}\cdot S(ABM)}+\frac{S(AC'M')}{\frac{BC}{CM}\cdot S(ACM)}\\ &=&\frac{BM}{BC}\cdot\frac{AB'\cdot AM'}{AB\cdot AM}+\frac{CM}{BC}\cdot\frac{AC'\cdot AM'}{AC\cdot AM}\end{eqnarray*} $\Longrightarrow BC\cdot\frac{AM}{AM'}=BM\cdot\frac{AC}{AC'}+CM\cdot\frac{AB}{AB'}$

Thanks. But also, can you tell why $A,K,S,L$ form a harmonic division (in Virgil's lemma)...

The division is harmonic because if two points are conjugate with respect to a circle (lie on one another's polars), then the chord joining those two points cuts the circle at a pair of points harmonically conjugate to them. The polar of A is EF, on which S lies, so A and S are conjugate and Virgil's result follows.

Ignore this post.

The problem can be solved using almost exclusively harmonic ratios. Let $D,E,F$ be the points of tangency of the incircle to the triangle, and $I$ the incenter. Now if $T$ is the point on $BC$ at infinity, then $(T,M;B,C)=-1$, so by projecting onto $EF$ from $A$, we have that $(R,Z;E,F)=-1$, where $R$ is the intersection of the line through $A$ parallel to $BC$ and $EF$, $Z$ is the intersection of $AM$ and $EF$. This means that $Z$ lies on the pole of $R$. But $Z$ also lies on the pole of $A$, so it is in fact the pole of $AR$, which makes $IZ\bot BC$. $X,Y$ are thus reflections of $K,L$ in $IZ$, so $X,Y,Z$ are collinear. Since $Z$ lies on the polar of $A$, we know that $(A,Z;K,L)=-1$. Projecting from $T$ onto $XY$ yields that $(W,Z;X,Y)=-1$, where $W$ is on $XY$ and $AR$. Finally, project these four harmonic points onto $BC$ from $A$ to show that $(T,M;P,Q)=-1$, which makes $M$ the midpoint of $PQ$.

It is well known that $EF, AL, DI$ are concurrent, where the incircle is tangent to $BC$ at $D$, $CA$ at $E$, $AB$ at $F$. Let $Z$ be this point of concurrence. $KXLY$ is an isosceles trapezium so $KL, XY, DI$ are concurrent. So $X,Y,Z$ is a straight line. Let $W$ be the point on $AX$ such that $YW || BC$ and $V$ be the point on $AY$ such that $VX || BC$. If $WZ$ intersects $AY$ at $U$ and $XU$ intersects $AM$ at $T$, $(A,T,Z,M)$ is harmonic. It is well known that $(A,K,Z,M)$ is harmonic, thus $K=T$. So $U=V$. By Ceva in $\triangle AYW$ we obtain that $YL=LW$ so $BP=CQ$.

[asy] import olympiad; size(300); pen pathpen = black + linewidth(0.7); pen pointpen = black; pen s = fontsize(8); pair B = (0,0), C = (5,0), A = (1,4); path w = incircle(A, B, C); pair I = incenter(A, B, C); pair E = intersectionpoint(w, A--C); pair F = intersectionpoint(w, A--B); pair D = intersectionpoint(w, B--C); pair M = midpoint(B--C); pair R = extension(E, F, D, I); pair K = intersectionpoint(w, A--M); pair L = intersectionpoint(R--M, w); pair a = foot(K, D, R); pair b = foot(L, D, R); pair X = 2*a - K; pair Y = 2*b - L; pair P = extension(A, Y, B, C); pair Q = extension(A, X, B, C); draw(A--B--C--A); draw(w, linetype("1 4")+linewidth(0.7)); draw(A--M); draw(E--F); draw(D--R); draw(F--K--X); draw(A--P); draw(A--Q); draw(L--Y--X); label("$A$", A, NW, s); label("$B$", B, SW, s); label("$C$", C, SE, s); label("$Q$", P, S, s); label("$P$", Q, S, s); label("$Y$", Y, SW, s); label("$L$", L, SE, s); label("$F$", F, NW, s); label("$E$", E, NE, s); label("$K$", K, NW, s); label("$D$", D, S, s); label("$M$", M, S, s); label("$X$", X, NE, s); label("$R$", R, SE, s); [/asy] Let $D$, $E$ and $F$ be the points of tangency of the incircle to $BC$, $AC$, and $AB$, respectively, and let $I$ be the incenter of $\triangle ABC$. By a well-known lemma, $EF$, $DI$, and $AM$ concur at a point $R$, so $(A, R; K, L) = -1$. Now, \begin{align*} \frac{AQ}{AP} = \frac{\sin \angle APQ}{\sin \angle AQP} = \frac{\sin \angle AYL}{\sin \angle AXK} = \frac{AX}{AY}\frac{AK}{AL} = \frac{AX}{AY}\frac{LR}{KR} = \frac{AX}{AY}\frac{YR}{XR} = \frac{\sin \angle PAM}{\sin \angle QAM} = \frac{PM}{QM} \frac{AQ}{AP}, \end{align*} implying that $BP = BM + PM = CM + QM = CQ$, as desired.

Let $D$, $E$, $F$ be the tangency points of $\gamma$ to $BC$, $AC$, and $AB$, respectively. It is well-known that $AM$, $EF$, and the diameter of $\gamma$ through $D$ concur at a point $Z$. Then it is clear that $(AZ; KL)$ is a harmonic bundle. Then we reflect over line $DX$ and get that $(A'Z; XY)$ is a harmonic bundle. We finish by projecting through $A$ onto line $BC$. Since $AA' \parallel BC$, we have that $(\infty M; PQ)$ is a harmonic bundle. This implies that $M$ is the midpoint of $PQ$ which means we must have $BP = CQ$, as desired.

Denote $B_1 \neq X$ and $C_1 \neq Y$ the intersection of $\omega$ with $AX$ and $AY$,respectively and the line parallel to $BC$through $Z$ intersects $\omega$ at $P_1$ and $Q_1$, where $P_1$ and $B$ is on the same side of $DI$. Also let $M$ be midpoint of $BC$ Note that, $BP = CQ$ is equivalent to $M$ is midpoint $PQ$ It is well known that $DI,EF$ and $AL$ is concurrented, say at $Z$. From La Haire theorem, we get that $B_1C_1, XY$ and $EF$ concurrent at $Z$. From Butterfly theorem, we get that $P_1Z = Q_1Z$. Since $P_1Q_1 \parallel BC$ and $A,Z,M$ collinear, we can conclude that $M$ is midpoint of $PQ$, as desired.

Let $DEF$ be the intouch triangle, let $EF\cap AM=G$ and let $AX\cap LY=N$. It's well known that $DI$, $AM$ and $EF$ are concurrent, so $D,I,G$ are collinear. It's clear that $DI$ is the perpendicular bisector of $XK$ and $YL$, so $X,Y,G$ are collinear. Then\begin{align*}\{P,Q;M,\infty \} & \underset{A}{=}\{N,Y;L,\infty \} \\ & \underset{X}{=}\{A,G;L,K\} \\ & \underset{E}{=}\{E,F;L,K\} \\ & =-1, \end{align*}so $M$ is the midpoint of $PQ$, which implies the desired result.

oops me when i learn ddit Claim: Let the tangents to $X_1$ and $Y_1$ to $\gamma$ meet at point $Z.$ Then $AZ \parallel BC.$ Proof: Let $I$ be the center of $\gamma,$ and let $D$ be the tangency point of $\gamma$ with $BC.$ By reflecting across $ID,$ it suffices to show that if the tangents to $\gamma$ at $X$ and $Y$ meet at point $W,$ then $AW \parallel BC.$ Let $S = EF \cap AM;$ then it is well-known that $DS \perp BC,$ but by polars $DS \perp AW$ as well. Thus $AW \parallel BC,$ as claimed. Now, applying DDIT, if $\gamma$ touches $AC,AB$ at points $E,F,$ respectively, there exists an involution swapping pairs $(AE,AF), (AZ, AZ), (AX_1, AX_2).$ Projecting onto $BC,$ we see that there exists an involution swapping $(C,B), (\infty, \infty),$ and $(P, Q).$ However, this is just the involution reflecting across the perpendicular bisector of $BC,$ and so $BP = CQ,$ as desired.

It suffices to show that $M$ bisects segment $\overline{QP}$. Let $Z = KX \cap AQ$, and $D, E, F$ denote the contact points of $\gamma$ with sides $\overline{BC}, \overline{CA}, \overline{AB}$, respectively. Before proceeding we introduce a lemma: Lemma: Let $\triangle ABC$ be a triangle with incenter $I$ and contact points $D, E, F$ on $\overline{BC}, \overline{CA},$ and $\overline{AB}$, respectively. If $M$ is the midpoint of $BC$, then $DI, AM,$ and $EF$ concur.

Now using this lemma in our context of $\overline{EF}, \overline{KL}, \overline{DI}$, we subsequently find that $X, G, Y$ are collinear, where $G$ is the concurrency point of $\overline{EF}, \overline{KL}, \overline{DI}$. Hence \[(Q, M; P, P_{\infty}) \overset{A}{\underset{KX}{\longrightarrow}} (Z, K; X, P_{\infty}) \overset{Y}{\underset{AM}{\longrightarrow}} (A, K; G, L) = -1, \]where the final equality follows from the lemma. $\blacksquare$

Let $J=KL \cap XY$, let $N$ be the point on $XY$ such that $AN \parallel BC$, let $D$, $E$ and $F$ be the intouch points on $BC$, $AC$, and $AB$ respectively. Now $(PQ ; M \infty) \stackrel{A}= (YX; JN) \stackrel{\text{rflct}}= (LK; JA)$. By EGMO lemma 4.17 $FJE$ are collinear so then $(LK;JA) \stackrel{\gamma}=(LK; EF)=-1$. Therefore $-1=(PQ; M \infty)$ and we can conclude $M$ is the midpoint of $PQ$ by midpoints and parallel lines.

Let $P_{\infty}$ be point at infinity on line $BC,$ and $D,E,F$ be the tangency points of the incircle. We have $$-1= (B,C;M,P_{\infty}) \overset{A}{=} (F,E;EF\cap AM, EF \cap AP_{\infty}).$$Let $T=EF\cap AM$. \textit{Claim:} $EF, AM=KL, XY$, and $DI$ are concurrent at $T$. \textit{Proof:} $EF, AM, DI$ are concurrent follows from homothethy and simson lines. It can alternativey be proven using homothethy and harmonic bundles. Since $XY$ is the reflection of $KL$ across $DI,$ we know that it is also concurrent. \textit{Claim:} $XE \cap FY, A, P_{\infty}$ are collinear. \textit{Proof:} It's easy to see $polar(A)=EF$ and $polar(P_{\infty}) = DI.$ By Brocard, $T \in polar(XE \cap FY).$ Since $DI \cap EF=T$, by duality of poles and polars (concurrency and collinearity), we are done. Taking perspecivity at $XE \cap FY \cap AP_{\infty},$ we get $$-1=(F,E;T, EF \cap AP_{\infty})=(Y, X; T, XY \cap AP_{\infty}) \overset{A}{=} (Q, P; M, P_{\infty}),$$which implies $M$ is the midpoint of $PQ,$ and the desired follows.

Let $\triangle DEF$ be the intouch triangle of $\triangle ABC$, and denote $R$ as the concurrency of $\overline{AM}$, $\overline{EF}$, and $\overline{DI}$. Moreover, let $S$ be the point on $\overline{XY}$ such that $\overline{AS} \parallel \overline{BC}$. We have \[-1 = (A,R;G,L) \overset{\infty}{=} (S,R;X,Y) \overset{A}{=} (\infty,M,P,Q),\] so $M$ is the midpoint of $\overline{PQ}$. This is equivalent to the desired conclusion, so we are done. $\square$

Let $\Delta DEF$ be the inotuch triangle of $\Delta ABC$, Let $DI \cap EF=G$. It is well known that $\overline{A-K-G-L-M}$ are collinear. Since $KY || XL || BC$, this implies $KXLY$ is an isoceles trapezoid and that $DI$ is the perpendicular bisector of $KY$ and $LX$.Since $G$ is the intersection of the perpendicular bisector of the parallel sides with one of the diagonals of a cyclic isoceles trapezoid, it follows that the other diagonal must pass through it as well, hence $\overline{X-G-Y}$ are collinear. Let $P_ \infty$ be the point at infinity along $BC$ and let the line through $A$ parallel to $BC$ intersect $XY$ at $T$. Also notice that $(K,L;E,F) \stackrel{E/F}{=} (K,L;A,G) =-1$. Now, $$(P,Q;M,P_\infty) \stackrel{A}{=} (X,Y;G,T) \stackrel{Reflection \ over \ DI}{=} (L,K;G,A)=-1$$It follows that $M$ is the midpoint of $PQ$ hence we are done.

I've given up on this problem twice so I kind of remember the solution but like whatever lol. Define $D,E,F$ as the intouch points. Construct $A'$, the reflection of $A$ over $ID$. Construct $Z=KL\cap XY$. Therefore, it also lies on $ID$, so by the ``an incircle concurrency'' lemma, it also lies on $EF$. Then \[-1=(EF;KL)\stackrel{F}{=}(ZA;KL)=(ZA';XY)\stackrel{A}{=}(M\infty_{BC};PQ),\]as desired.

Apply DDIT on $KXLY$ through $A$, with helper line $\ell=\overline{BC}$. Then there exists an involution on $\ell$ swapping $Q$ and $P$, $M$ and $M$, and $\infty$ with $A(KY\cap XL)\cap \ell$. Note that $Z=KY\cap XL$ is the Miquel point of $KXLY$, so $D=KL\cap XY$ is the inverse of $Z$ wrt the incircle. We want to show that $AZ\parallel BC$, which is equivalent to $A$ being on the polar of $D$. This is equivalent to $D$ being on the polar of $A$, which is the $A$-intouch chord. This is exactly the "an incircle concurrency" lemma. $\blacksquare$

What's projective? By the two homotheties at $A$ mapping either $\overline{KX}$ or $\overline{LY}$ to $\overline{BC}$, there exist two circles through $M$ tangent to $\overline{AB}$ and $\overline{AC}$ intersecting $\overline{BMC}$ again at $P$ and $Q$. Let one of these circles be $\omega$, and suppose it is tangent to $\overline{AB}$ at $D_1$ and $\overline{AC}$ at $E_1$. We will resolve the case when $D_1$ is on segment $AB$; other cases are analogous. Let $P'$ be the reflection of $P$ over $M$. We will construct a circle through $M$ and $P'$ tangent to $\overline{AB}$ and $\overline{AC}$. Choose $D_2$ such that $BD_2=CE_1$ and $B$ lies on segment $AD_2$, and choose $E_2$ such that $CE_2=BD_1$ and $C$ lies on segment $AE_2$. Notice that since $BD_2^2=CE_1^2=CM \cdot CP=BM \cdot BP'$, there exists a circle $\omega_1$ through $M$ and $P'$ tangent to $\overline{AB}$ at $D_2$ since $CE_2^2=BD_1^2=BM \cdot BP=CM \cdot CP'$, there exists a circle $\omega_2$ through $M$ and $P'$ tangent to $\overline{AC}$ at $E_2$ since $AD_1+D_1B+BD_2=AE_1+E_1C+CE_2$, there exists a circle $\omega_3$ tangent to $\overline{AB}$ at $D_2$ and $\overline{AC}$ at $E_2$. If any two of $\omega_1$, $\omega_2$, and $\omega_3$ coincide, then they all coincide and we are done. Otherwise, radical axis on the three circles implies lines $AB$, $AC$, and $BC$ concur, a contradiction. $\blacksquare$

Let $D, E, F$ denote the point of tangency of $\gamma$ with sides $BC, CA, AB$ respectively. Lemma: Let $I$ denote the incenter of $\triangle ABC$, with $M$ being the midpoint of $AB$. Then: $AM, EF, DI$ are concurrent. Proof: Let $X = DI \cap EF$. Let $B'C'$ be a line passing though $X$ with $B'C' \parallel BC$. Notice that, the pedal triangle of point $I$ onto $\triangle AB'C'$ is degenerate which implies $I$ lies on $\triangle AB'C'$. Since $AI$ is the angle bisector of $\angle B'AC'$, and $X$ is the foot of perpendicular of $I$ onto $B'C'$, we have: $X$ to be the midpoint of $BC$. Taking homothety gives us the result. Note that, $KXLP$ is an isosceles trapezoid. Due to the above lemma, let $R$ denote the concurrency point of lines $AM, EF, DI$. Note that: $R$ lies on $XP$. Let $A'$ be a point such that: $AA' \parallel BC$ with $A', X, P$ collinear. Notice that, as we get $(A, A'), (K, X), (R, R), (P, L)$ are symmetric with-respect to line $DR$: $$-1=(AR;KL) = (A'R; XP).$$Taking perspective into line $BC$ from point $A$ gives: $$-1 = (A'R; XP) = (\infty M; QP).$$Therefore $M$ is the midpoint of $PQ$ and we are done.

It is well known that the polar of $AM\cap EF$ where $EF$ are the incircle touch points on $AB$ and $AC$ is the line through $A$ parallel to $BC$ and also that if $D$ is the last touch point that the perpendicular from $D$ goes through $AM\cap EF$. Now let $AM\cap EF$ be $J$. Clearly $XY$ passes through $J$ so projecting from $A$ proves the result.