Let $M$ and $N$ be two points on the side BC of a triangle $ABC$ such that $BM =MN = NC$. A line parallel to $AC$ meets the segments $AB, AM$ and $AN$ at the points $D, E$ and $F$ respectively. Prove that $EF = 3DE$
Source: 1999 Singapore TST 2.1
Tags: geometry, equal segments
Let $M$ and $N$ be two points on the side BC of a triangle $ABC$ such that $BM =MN = NC$. A line parallel to $AC$ meets the segments $AB, AM$ and $AN$ at the points $D, E$ and $F$ respectively. Prove that $EF = 3DE$