Points $A,B,C,D,E,F,G,H$ lie, in that order, on $\overline{AH}$, dividing it into seven segments, each of length $1$. Point $K$ is not on line $AH$. Point $L$ lies on $\overline{KD}$, and point $M$ lies on $\overline{KH}$. The line segments $\overline{LC}, \overline{ME},$ and $\overline{AK}$ are parallel. Find $LC/ME$.
Solution
Put $A$ at the origin, so $A,B,C,D,E,F,G,H$ are at $0,1,\dots,7$ on the line. Since $ME\parallel AK$, $\triangle HEM\sim\triangle HAK$, so $\dfrac{EM}{AK}=\dfrac{HE}{HA}=\dfrac37$. Since $LC\parallel AK$, $\triangle DCL\sim\triangle DAK$, so $\dfrac{CL}{AK}=\dfrac{DC}{DA}=\dfrac13$.
Hence $\dfrac{LC}{ME}=\dfrac{1/3}{3/7}=\boxed{\dfrac79}$.
