Problem

Source:

Tags: geometry, ratio



In the triangle $ABC$, let $D \in (BC)$, $E \in (AB)$, $EF \parallel BC$, $F \in (AC)$, $EG\parallel AD$, $G\in (BC)$ and $M,N$ be the midpoints of $(AD)$ and $(BC)$, respectively. Prove that: a) $\frac{EF}{BC}+\frac{EG}{AD}=1$ b) the midpoint of $[FG]$ lies on the line $ MN$.