Belarusian Mathematical Olympiad

Construct the segment $AK$. By condition, $T$ lies on $AK$. Let $R$ be the intersection point of lines $DT$ and $KM$ (see the figures). Let $AN=x$, $ND=y$, $RM=z$, $AT:TK=1:k$. By Thales' theorem, $k=TK:AT=y:x$ (since $NT\parallel DK$). Since $NMKD$ is a parallelogram, we have $MK=ND=y$, and then $$z:y=RM:MK=[TM\parallel DK]=RT:TD=[TC\parallel RK]=KC:CD=$$ $$=[TC\parallel AD]=KT:TA=k.(1)$$ So, $y=kx$, $z=ky=k^{2}x$. Let $P_1$ be the point of intersection of $AM$ and $RD$ (Fig. 1).





Construct a line passing through $A$ parallel to $AM$, and let $Q_1$ be the point of intersection of this line and the line $MK$. By Thales' theorem, $$\begin{array}{c}R{P}_1:P_{1}D=RM:M{Q}_1=[AM{Q}_{1}D\text{ is a parallelogram,}\\ M{Q}_1=AD=x+y]=z:(x+y)=k^{2}x:(x+kx)=\frac{k^2}{1+k}.\end{array}$$ Therefore, $$R{P}_1=\frac{k^2}{1+k}{P}_{1}D=\frac{k^2}{1+k}(RD-R{P}_1)\Rightarrow R{P}_1=\frac{k^2}{1+k+k^2}RD.(2)$$

Let $P_2$ be the point of intersection of the segments $BK$ and $RD$ (Fig. 2). Construct a line passing through $T$ parallel to $BK$, and let $Q_2$ be the point of intersection of this line with the line $MK$. By Thales' theorem, $$R{P}_2:P_{2}T=RK:K{Q}_2=[BK{Q}_{2}T,BTNA,\text{NMKD is a parallelogram,}]$$ $$K{Q}_2=BT=AN=x,MK=ND=y]=(z+y):x=(k^2+k).$$ Therefore, $$R{P}_2=(k^2+k)P_{2}T=k(k+1)(RT-R{P}_2)⟹$$ $$R{P}_2=\frac{k(k+1)}{k^2+k+1}RT=[RT=kTD=k(RD-RT)⟹$$ $$RT=\frac{k}{k+1}RD]=\frac{k^2}{k^2+k+1}RD.$$ Comparing this result with (2), we see that $P_1$ coincides with $P_2$, and so they coincide with $P$, which gives the required statement.