Given is the triangle $ABC$, with $| AB | + | AC | = 3 \cdot | BC | $. Let's denote $D, E$ also points that $BCDA$ and $CBEA$ are parallelograms. On the sides $AC$ and $AB$ sides, $F$ and $G$ are selected respectively so that $| AF | = | AG | = | BC |$. Prove that the lines $DF$ and $EG$ intersect at the line segment $BC$
Problem
Source: Czech-Polish-Slovak Junior Match 2017, individual p2 CPSJ
Tags: geometry, concurrency, concurrent