Question 132
A point $D$ chosen on the side $BC$ of a triangle $ABC$ such that $AD = 2BD$. Let $E$ and $F$ be the foot of the perpendicular line from $D$ to the line $AB$ and the line $AC$, respectively. A line through $D$ perpendicular to the line $BC$ meets with $EF$ at $M$. A line through $M$ and parallel to $BC$ meets with $AB$ at $K$ and meets with side $AC$ at $L$. Prove that $DL = 2BK$.