Hard problem I think. No synthetic solution after 2 hours of work, so I just decided to bash it. Turns out that it's quite simple to bash after few synthetic ideas. Perhaps it's because many properties of known configurations are hidden inside a square, so there's no trigger to remind you which one is important. Consider the following Lemma. Let $ABC$ be a triangle with $\angle C=45^{\circ}$. Tangent from $A, B$ to the circumcircle of triangle $ABC$ meet $CB$ and $CA$ at $K, L$ respectively. Let $M, N$ be projections of $K, L$ onto $AB$, respectively. Then, lines $KN$ and $ML$ meet on the $C$-median of $\triangle ABC$. (Proof) Let $R$ be midpoint of $AB$. Note that $$\boxed{\frac{d(R, CK)}{d(R, CL)}=\frac{\sin A}{\sin B}=\frac{CB}{CA}}.$$Observe that $$\boxed{\frac{d(N, KL)}{d(M, KL)}=\frac{NL}{KM}=\frac{BL}{KA} \cdot \frac{CA}{CB}}.$$Finally, we show that $$\boxed{\frac{d(M, CL)}{d(N, CK)}=\frac{KA}{LB}}.$$Indeed, $$\frac{d(M, CL)}{d(N, CK)}=\frac{MB \cos B}{NA \cos A}=\frac{KA \cdot \cos A \cos B+AB \cos B}{LB \cdot \cos A \cos B+AB \cos A},$$hence it suffices to show that $$\frac{KA}{LB}=\frac{\cos B}{\cos A}.$$By sine rule in $\triangle KAB$ and $\triangle LAB$ we conclude that $$\frac{KA}{LC}=\frac{\sin (A-45^{\circ})}{\sin (B-45^{\circ})}.$$Together with $A+B=135^{\circ}$ we conclude that $$\sin (2A-45^{\circ})=\sin(2B-45^{\circ}) \implies \frac{\sin (A-45^{\circ})}{\sin (B-45^{\circ})}=\frac{\cos B}{\cos A},$$as desired. Multiply the boxed equations and apply distance form of Ceva's Theorem to conclude the lemma. $\blacksquare$ Back to our problem, let $PA$ and $PB$ meet $CD$ at $A', B'$. Note that $\angle KPB'=\angle KDB'=45^{\circ}$ so $K, B', P, D$ are concyclic. Similarly $L, A', P, C$ are concyclic. To conclude, apply the lemma on $\triangle PA'B'$. $\blacksquare$ Note. $d(X, \ell)$ is the orthogonal distance between point $X$ and line $\ell$.

Official solution http://geometry.ru/olimp/2017/final_sol_en.pdf

Let $O$ be the center of $(ABCD)$ and let $X$ be the foot from $Q$ to $CD$. Let $R$ be the radius of $(ABCD)$. Lemma. $P$, $X$, $O$ are collinear. Proof. Note that we have circles $(OLNPD)$ and $(OKMPC)$; their radical axis is just $OP$. We wish to show that $X$ lies on this radical axis. To do this, we note that the powers of $X$ are $XD\cdot XN$ and $XC \cdot XM$. In other words, we only need to prove that $\frac{XN}{XM} = \frac{XC}{XD}$. But we note that \[ \frac{XN}{XM} = \frac{LN}{KM} = \frac{NC}{MD} = \frac{XN + NC}{XM + MD} = \frac{XC}{XD}, \]as desired. $\blacksquare$ Let $S$ be the midpoint of $AB$. It suffices to show that triangles $PQX$ and $PSO$ are similar. Since $SO$ and $QX$ are parallel, it suffices to show that $\frac{PX}{XQ} = \frac{PO}{OS}$. But note that $\frac{PO}{OS} = \sqrt{2}$; thus our mission is to show that $\frac{PX}{XQ} = \sqrt{2}$, or equivalently \[OX^2 = 2 (\frac{R}{\sqrt{2}} - QX)^2.\] We will introduce variables. Let $a = DM$ and $b = CN$; note that $KM = a$, $KD = a\sqrt{2}$, and $OK = R - a\sqrt{2}$. Similar relations hold for $b$. Let's try to find a relationship between $a$ and $b$. Note that $\frac{OL}{PD} = \frac{OB}{BP}$; thus $OL = R \cdot \frac{PD}{PB}$. Since $PA$ bisects $\angle BPD$, we have $OL = R \cdot \frac{KD}{KB}$. This implies that \[R - b\sqrt{2} = (R)(\frac{a\sqrt{2}}{2R - a\sqrt{2}}).\]A little bit of expansion yields \[R^2 - (a+b)\sqrt{2}R + ab = 0.\] It's easy to calculate \[XQ = \frac{1}{\frac{1}{KM} + \frac{1}{LN}} = \frac{ab}{a+b}.\]Also, we find easily \[OX^2 = R^2(\frac{a^2+b^2}{(a+b)^2}) = \frac{a^2+b^2}{a+b}\sqrt{2}R - \frac{ab(a^2+b^2)}{(a+b)^2}\]using Stewart's Theorem on triangle $OCD$. From here it's very easy to verify that \[OX^2 = 2(\frac{R}{\sqrt{2}} - QX)^2,\]as desired. $\blacksquare$

Thank you for your interest. I am very happy when my problem was appeared in the greatest contest of geometry - Sharygin olympiad !

I would like to propose another bisecting problem in this configuration of square. I only note that, all this configuration was created by myself when I study my own problem for contest of class 10 HSGS. Extension on configuration. Let $ABCD$ be a square, and let $P$ be a point on the minor arc $CD$ of its circumcircle. The lines $PA,$ $PB$ meet the diagonals $BD,$ $AC$ at points $M,$ $N$ and meet side $CD$ at $S,$ $T$ respectively. $K,$ $L$ are the orthogonal projection of $M,$ $N$ on $AD,$ $BC$ respectively. $KS$ cuts $TL$ at $Q.$ Prove that $PQ$ bisects the segment $AB.$ I also would like to present the nice generalization for this problem by my friend andria in link.

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buratinogigle wrote: I would like to propose another bisecting problem in this configuration of square. I only note that, all this configuration was created by myself when I study my own problem for contest of class 10 HSGS. Extension on configuration. Let $ABCD$ be a square, and let $P$ be a point on the minor arc $CD$ of its circumcircle. The lines $PA,$ $PB$ meet the diagonals $BD,$ $AC$ at points $M,$ $N$ and meet side $CD$ at $S,$ $T$ respectively. $K,$ $L$ are the orthogonal projection of $M,$ $N$ on $AD,$ $BC$ respectively. $KS$ cuts $TL$ at $Q.$ Prove that $PQ$ bisects the segment $AB.$ I also would like to present the nice generalization for this problem by my friend andria in link. Dear Mr Hung, I think this problem and the problem in Sharygin Olympiad 2017 are nice problems in your geometry collection which are created by the model with a point on the circumcircle of a square. Here is my solution for problen in post #6.

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Dear Mathlinkers, have also a look at http://jl.ayme.pagesperso-orange.fr/Docs/Sequence%201.pdf p. 41... Sincerely Jean-Louis

anantmudgal09 wrote: Let $ABCD$ be a square, and let $P$ be a point on the minor arc $CD$ of its circumcircle. The lines $PA, PB$ meet the diagonals $BD, AC$ at points $K, L$ respectively. The points $M, N$ are the projections of $K, L$ respectively to $CD$, and $Q$ is the common point of lines $KN$ and $ML$. Prove that $PQ$ bisects the segment $AB$. Nice problem. Probably hard for 8 though and not at all coordinate bashable. Let the parallel to $AB$ through $Q$ meet $KM, LN$ at $X, Y$ respectively. Let $E$ be the foot of perpendicular from $Q$ to $AB$. Let $KP \cap CD = U, LP \cap CD = V$. Note that $M, N$ are circumcentre of $\triangle KDP, \triangle LCP$ respectively. So, $\angle MPN = 90^{\circ}$. Also, $MK = MV = MP$ and $NL = NU = NP$. Also, $$\frac{QX}{NU} = \frac{KQ}{KN} = \frac{MQ}{ML} = \frac{QE}{LN} = \frac{QE}{NU}$$So, $QX = QE$. Similarly, $QY = QE$. Thus, $QX = QY$. From this the claim easily follows. $\blacksquare$.

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Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf p. 25... Sincerely Jean-Louis