60th Belarusian Mathematical Olympiad

Let $R$ be the midpoint of $BD$, $M$ be the midpoint of $AD$, $P$ be the point of intersection of $AC$ and $BD$. We show that $LD=2LP$.



By condition, $BP:PD=1:3$. So $BP=PR=0.5RD$, hence $LP$ is the median of the triangle $BLP$, and $LR$ is the median of the triangle $BLD$.

Using the formula for the median length, we obtain $$4L{P}^2=2L{B}^2+2L{R}^2-B{R}^2,4L{R}^2=2L{B}^2+2L{D}^2-B{D}^2.$$ Thus $$\begin{array}{rl}4L{P}^2& =2L{B}^2+2L{R}^2-B{R}^2=2L{B}^2+0.5(2L{B}^2+2L{D}^2-B{D}^2)-B{R}^2=\\ & =3L{B}^2+L{D}^2-0.5(2BR{)}^2-B{R}^2=L{D}^2-3(B{L}^2-B{R}^2)=[BL=BR]=L{D}^2.\end{array}$$ Similarly, $LA=2LP$. Therefore, $LA=LD$, i.e. $L$ lies on the perpendicular bisector of $AD$. Similarly, $K$ lies on the perpendicular bisector of $AD$. Hence the line $LK$ meets $AD$ at its midpoint.