Answer:
[tex]\dfrac{DF}{FC}=\dfrac{1}{1}[/tex]
Step-by-step explanation:
Consider triangles ABE and CFE. Since ABCD is parallelogram, lines AB and CD are parallel. Thus, angles ABE and EFC and BAE and ECF are congruent as alternate interior angles. Angles BEA and FEC are congruent as vertical angles. Hence, triangles ABE and CFE are similar by AA theorem.
Similar triangles have proportional sides:
[tex]\dfrac{AE}{EC}=\dfrac{AB}{CF}=\dfrac{BE}{EF}.[/tex]
From this ratio, you have
[tex]\dfrac{AB}{CF}=\dfrac{2}{1}.[/tex]
In parallelogram ABCD, AB=CD, hence
[tex]\dfrac{CD}{CF}=\dfrac{2}{1}\Rightarrow CD=2CF.[/tex]
Since CD=CF+DF and CD=2CF, you have 2CF=CF+DF, DF=CF. Therefore,
[tex]\dfrac{DF}{FC}=\dfrac{1}{1}.[/tex]
