Call $ PQ\cap AD=T$ Use Menelaus theorem for $ 2$ triangle $ ABD,ACD$ to prove $ \frac{AP}{PC}=\frac{BQ}{QD}$. Then $ AP.PC=BQ.QD$ mean $ AP=BQ,PC=QD,AC=BD$ Then OK!

Misollar unchalik qiyin emaskanu, lekin Uzbekistan faqatgina bitta bronze medal olgani chatoq bo'lipdida! I think my solutin isn't nice. Let $ AC\cap BD=O$.Sice points$ K,P,Q$ are collinear using Menelaus theorem we have $ \frac{BQ}{AP}=\frac{QO}{OP}$, because $ AK=KB$.Similarly $ \frac{PC}{QD}=\frac{OP}{OQ}$.Then condition of the problem states that $ QO=OP$.Then $ \angle LKM=\angle OPQ=\angle OQP=\angle LMK$.Hence $ KLMN$ is a rhombus, i.e $ LN\perp KM$.Then $ \angle ORS=90^\circ-\angle OPQ=\angle OSR$.Hence $ OR=OS$.(1) Since points $ R,S,N$ are collinear using Menelaus theorem we have $ \frac{AR}{SD}=\frac{OR}{OS}$ (2) , because $ AN=ND$.Similarly$ \frac{BS}{RC}=\frac{SO}{OR}$ (3).From (1),(2) and (3) we have $ 1=\frac{AR}{SD}=\frac{BS}{RC}$ or $ AR\cdot RC=BS\cdot SD$ and we are done.

Ibrogimov wrote: Misollar unchalik qiyin emaskanu, lekin Uzbekistan faqatgina bitta bronze medal olgani chatoq bo'lipdida! Please,write in english. My solution to this problem is very short.Hope some of you will find it.

In our team one boy solved this by trigonometric methods.[/quote]

let $U,V$ be the intersections of AB and CD;AD and BC. by Menelaus,$AP*PC=\frac{AP*PC}{AC^2}*AC^2=\frac{CM*UK*MU*KA}{CM*UK+MU*KA)^2}*AC^2=\frac{DM*MU*UK*KB}{(DM*UK+MU*KB)^2}*BD^2=BQ*BD$ hence $AC=BD$ analagously,$AR*RC=BS*SD\leftrightarrow AC=BD$.

Here's a bashy way to get to the magical $AC = BD$. Let $X$ and $Y$ be the midpoints of segments $AC$ and $BD$, respectively, and $O = \overline{AC} \cap \overline{BD}$. Let $G$ be the centroid of $ABCD$, or otherwise $G = \overline{XY} \cap \overline{KM} \cap \overline{NL}$. WLOG, we can assume that $G$ lies inside $\triangle BOC$. Let $Z = \overline{GX} \cap \overline{MN}$ and $T = \overline{GY} \cap \overline{NK}$. The idea of this setup is to interpret $AP\cdot PC$, $BQ \cdot QD$, $AR \cdot RC$, and $BS\cdot SD$ through the midpoints $X$ and $Y$ as the powers of the points $P, Q, R, S$ with respect to the circles with diameters $AC$ and $BD$. That is, $AP \cdot PC = AX^2 - XP^2$, $BQ\cdot QD = BY^2 - YQ^2$, $AR \cdot RC = AX^2 - XR^2$, and $BS \cdot SD = BY^2 - YS^2$. Given that $AP \cdot PC = BQ \cdot QD$, proving $AR\cdot RC = BS \cdot SD$ is equivalent to showing that $XP^2 - XR^2 = YQ^2 - YS^2$. For that reason, we introduced points $Z$ and $T$, and using Thales's theorem repeatedly, we get that: \begin{align*} &XP^2 - XR^2 = YQ^2 - YS^2 \\ &\iff \left(\frac{GR}{GN}\right)^2 \cdot (ZM^2 - ZN^2) = \left(\frac{GY}{GT}\right)^2 \cdot (TK^2 - TN^2) \\ &\iff \left(\frac{\text{dist}(G, \overline{AC})}{\text{dist}(G, \overline{MN})}\right)^2 \cdot (ZM^2 - ZN^2) = \left(\frac{\text{dist}(G, \overline{BD})}{\text{dist}(G, \overline{NK})}\right)^2 \cdot (TK^2 - TN^2) \end{align*}However, we have that \[\text{dist}(G, \overline{AC}) = \frac{1}{2}\left(\text{dist}(B, \overline{AC}) - \text{dist}(D, \overline{AC})\right)\quad \text{and} \quad \text{dist}(G, \overline{MN}) = \text{dist}(G, \overline{AC}) + \frac{1}{2}\text{dist}(D, \overline{AC}) = \frac{1}{2}\text{dist}(B, \overline{AC})\]Similar expressions can be obtained for $\text{dist}(G, \overline{BD})$ and $\text{dist}(G, \overline{NK})$, whence \begin{align*} &XP^2 - XR^2 = YQ^2 - YS^2 \\ &\iff \left(\frac{\text{dist}(B, \overline{AC}) - \text{dist}(D, \overline{AC})}{\text{dist}(B, \overline{AC})}\right)^2 \cdot (ZM^2 - ZN^2) = \left(\frac{\text{dist}(C, \overline{BD}) - \text{dist}(A, \overline{BD})}{\text{dist}(C, \overline{BD})}\right)^2 \cdot (TK^2 - TN^2)\\ &\iff \left(1 - \frac{DO}{BO}\right)^2 \cdot (ZM^2 - ZN^2) = \left(1 - \frac{AO}{CO}\right)^2 \cdot (TK^2 - TN^2) \end{align*}Now, by Menelaus's theorem for $\triangle MNL$ and the line $Z, G, T$, we have that $\frac{KT}{TN} \cdot \frac{NZ}{ZM} = 1$ as $G$ is the midpoint of $MK$. Substituting in $r = \frac{TK}{TN}$ for brevity implies $\frac{ZM}{ZN} = r$ and so we get: \begin{align*} &XP^2 - XR^2 = YQ^2 - YS^2 \\ &\iff \left(1 - \frac{DO}{BO}\right)^2 \cdot (ZM^2 - ZN^2) = \left(1 - \frac{AO}{CO}\right)^2 \cdot (TK^2 - TN^2)\\ &\iff \left(1 - \frac{DO}{BO}\right)^2 \cdot \frac{r^2-1}{(r+1)^2}\cdot MN^2= \left(1 - \frac{AO}{CO}\right)^2 \cdot \frac{r^2-1}{(1-r)^2} \cdot NK^2 \\ &\iff \frac{\left(1 - \frac{DO}{BO}\right)^2}{\left(1 - \frac{AO}{CO}\right)^2} \cdot \frac{AC^2}{BD^2} = \left(\frac{1+r}{1-r}\right)^2 \end{align*}The initial condition from the problem statement should imply the last line, but \begin{align*} &AP\cdot PC = BQ \cdot QD\\ &\iff AC^2 - 4XP^2 = BD^2 - 4YQ^2 \\ &\iff AC^2 - BD^2 = 4\left(\frac{GR}{GN}\right)^2\cdot ZM^2 - 4\left(\frac{GY}{GT}\right)^2 \cdot TK^2 \\ &\iff AC^2 - BD^2 = 4\left(1 - \frac{DO}{BO}\right)^2 \cdot \frac{r^2}{(1+r)^2} \cdot MN^2 - 4 \left(1 - \frac{AO}{CO}\right)^2 \cdot \frac{r^2}{(1-r)^2} \cdot NK^2 \\ &\iff AC^2 - BD^2 = \left(1 - \frac{DO}{BO}\right)^2 \cdot \frac{r^2}{(1+r)^2} \cdot AC^2 - \left(1 - \frac{AO}{CO}\right)^2 \cdot \frac{r^2}{(1-r)^2} \cdot BD^2 \\ &\iff AC^2 \cdot \left(1 - \frac{r^2}{(1+r)^2}\cdot \left(1 - \frac{DO}{BO}\right)^2\right) = BD^2 \cdot \left(1 - \frac{r^2}{(1-r)^2}\cdot \left(1 - \frac{AO}{CO}\right)^2\right) \end{align*}Comparing this equation with the previous $ \frac{\left(1 - \frac{DO}{BO}\right)^2}{\left(1 - \frac{AO}{CO}\right)^2} \cdot \frac{AC^2}{BD^2} = \left(\frac{1+r}{1-r}\right)^2$ shows that the two are equivalent if and only if $\frac{AC}{BD} = 1$. Having conjectured this claim, it isn't that hard to prove. If $\overline{MK}\cap\overline{AD} = E$, then by Menelaus's theorem applied on $\triangle ACD$ and the line $E, M, P$ and $\triangle ABD$ and the line $E, Q, K$ implies that $\frac{AP}{PC} = \frac{BQ}{QD}$, which readily implies $AC = BD$ because $AP \cdot PC = BQ \cdot QD$.