12.1.5 Olympic geometry

Problem: On the side \( C D \) of the parallelogram \( A B C D \) points \( E_{1} \) and \( E_{2} \) are chosen so, that \( A B=B E_{1}=B E_{2} \). On the beam \( A E_{1} \) the point \( F_{1} \) is chosen so, that \( B E_{1} \| C F_{1} \), and on the beam \( A E_{2} \) the point \( F_{2} \) is chosen so, that \( B E_{2} \| C F_{2} \). Prove that \( D F_{1}=D F_{2} \).